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Mathematical induction
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92
Appl
ied
Mat
hem
atic
s 11
Man
y in
tere
stin
g an
d im
port
ant r
esul
ts in
mat
hem
atic
s ha
ve b
een
disc
over
ed b
y fir
st o
bser
ving
patt
ern
s in
som
e sp
ecif
ic c
ases
an
d th
en m
akin
g ge
ner
alis
atio
ns
from
th
e ob
serv
atio
ns,
th
at
is, b
y ap
plyi
ng
indu
ctiv
e re
ason
ing.
For
exa
mpl
e, it
is e
asy
to fo
rm a
con
ject
ure
abo
ut t
he
sum
of t
he
firs
t k
odd
posi
tive
inte
gers
.
1=
1
1 +
3=
4
1 +
3 +
5=
9
1 +
3 +
5 +
7=
16
1 +
3 +
5 +
7 +
9=
25
. . .
Th
e n
um
bers
1,
4, 1
6, a
nd
25 a
re t
he
squ
ares
of
the
firs
t fi
ve p
osit
ive
inte
gers
, an
d so
it
appe
ars
that
the
sum
of t
he fi
rst k
odd
posi
tive
inte
gers
is e
xact
ly k
2 . Sin
ce th
e kt
h od
d po
sitiv
e
inte
ger
may
be
wri
tten
as
(2k
– 1)
, th
e co
nje
ctu
re m
ay a
lso
be e
xpre
ssed
as
an e
quat
ion
:
1 +
3 +
5 +
... +
(2k
– 1)
= k
2
Alt
hou
gh t
his
par
ticu
lar
form
ula
is
valid
, it
is
impo
rtan
t fo
r yo
u t
o se
e th
at r
ecog
nis
ing
a
patt
ern
an
d th
en s
impl
y ju
mpi
ng
to t
he
con
clu
sion
th
at t
he
patt
ern
mu
st b
e tr
ue
for
all
valu
es o
f n
is n
ot a
logi
cally
val
id m
eth
od o
f pr
oof.
Th
ere
are
man
y ex
ampl
es in
wh
ich
a p
at-
tern
app
ears
to
be d
evel
opin
g fo
r sm
all v
alu
es o
f n
an
d th
en a
t so
me
poin
ts t
he
patt
ern
fai
ls.
On
e of
the
mos
t fam
ous
case
s of
this
was
the
con
ject
ure
by
the
Fren
ch m
ath
emat
icia
n P
ierr
e
de F
erm
at (
1601
- 1
655)
, wh
o sp
ecu
late
d th
at a
ll n
um
bers
of
the
form
F n=
22n
+ 1
, n=
0,1
,2, .
..
are
prim
e. F
or n
= 0
, 1, 2
, 3, a
nd
4, t
he
con
ject
ure
is t
rue.
Th
e fi
fth
Fer
mat
nu
mbe
r (F
5=
429
4967
297)
is
so g
reat
th
at i
t w
as d
iffi
cult
for
Fer
mat
to
dete
rmin
e w
het
her
it
was
pri
me
or n
ot.
How
ever
, an
oth
er w
ell-
know
n m
ath
emat
icia
n,
Leo
nh
ard
Eu
ler
(170
7 -
1783
) , l
ater
fou
nd
the
fact
oris
atio
n
F 5=
429
4967
297
= 6
41(6
7004
17)
wh
ich
pro
ved
that
F5
is n
ot p
rim
e, a
nd
ther
efor
e Fe
rmat
's co
nje
ctu
re w
as f
alse
.
Just
bec
ause
a r
ule
, pa
tter
n,
or f
orm
ula
see
ms
to w
ork
for
seve
ral
valu
es o
f n
, yo
u c
ann
ot
sim
ply
deci
de t
hat
it is
val
id f
or a
ll va
lues
of
n w
ith
out
goin
d th
rou
gh a
legi
tim
ate
proo
f.
93
Mat
hem
atic
al In
duct
ion
A.
MA
TH
EM
ATIC
AL I
ND
UC
TIO
N
Mat
hem
atic
al in
duct
ion
is b
ased
on
sim
ple
char
acte
rist
ics
of t
he
nat
ura
l nu
mbe
rs w
hic
h a
re
stat
ed in
th
e fo
llow
ing
theo
rem
.
Th
eo
rem
If I
Dis
a s
ubs
et o
f n
atu
ral n
um
bers
so
that
1.1
IDan
d
2.k
IDim
plie
s k
+ 1
ID
, th
en I
D=
IN
+
Pro
of
On
th
e co
ntr
ary,
that
ID
IN+,
then
th
ere
exis
ts a
t le
ast
one
elem
ent,
say
mIN
+
such
th
at
mID
. S
o th
is
impl
ies
that
(m
– 1)
ID
, an
d if
(m
– 1)
ID
, th
en
(m–
1) –
1 =
(m
– 2)
ID
. By
this
way
, we
can
eas
ily s
ee t
hat
,
(m–
3), (
m–
4), .
.. , 3
, 2, 1
will
not
be
an e
lem
ent
of I
D. B
ut
by in
duct
ion
th
eore
m, 1
ID
, so
we
con
clu
de t
he
this
is a
con
trad
icti
on. T
her
efor
e, I
D=
IN
+.
Indu
ctio
n T
heo
rem
Let
IN
+
be t
he
non
-em
pty
set
wit
h
1.1
IN+
.
2.Fo
r ea
ch n
IN+,
n*
IN+
cal
led
succ
esso
r of
n.
3.Fo
r ea
ch n
IN+, w
e h
ave
n*
1.
4.If
m, n
IN
+an
d m
* =
n*,
th
en m
= n
.
5.A
n s
ubs
et K
of I
N+
hav
ing
the
prop
erti
es
a.1
K
b.k*
K
wh
enev
er k
Kis
equ
al t
o IN
+
In o
rder
to
prov
e a
prop
osit
ion
by
mat
hem
atic
al i
ndu
ctio
n,
it i
s en
ough
to
sati
sfy
the
fifth
Pean
o A
xiom
. Th
ese
axio
ms
will
inve
stig
ated
in a
dvan
ced
alge
bra.
Axio
m
Indu
ctio
n t
heo
rem
is
use
ful
part
icu
larl
y fo
r pr
ovin
g pr
opos
itio
ns
invo
lvin
g th
e po
siti
ve
inte
gers
. To
form
ula
te t
he
follo
win
g pr
inci
ple
of m
ath
emat
ical
indu
ctio
n, i
t is
con
ven
ien
t to
let
Pn
be a
pro
posi
tion
wh
ere
n is
th
e po
siti
ve in
tege
r.
Co
nclu
sio
n
Let
aIN
+an
d P
nbe
a p
ropo
siti
on in
volv
ing
the
posi
tive
inte
ger
na.
If,
1.th
e pr
opos
itio
n is
tru
e fo
r n
= 0
, th
at is
Pa
is t
rue,
an
d
2.fo
r ev
ery
posi
tive
inte
ger
k, t
he
tru
th o
f P
kim
plie
s th
e tr
uth
of
Pk
+ 1, t
hen
Pn
mu
st b
e tr
ue
for
all p
osit
ive
inte
gers
n.
Pri
nci
ple
of M
ath
emat
ical
In
duct
ion
94
Appl
ied
Mat
hem
atic
s 11
The
set
Na
= {
n|
na}
whe
re a
IN+
is c
alle
d th
e tr
uth
set
of P
n. N
ote
that
, aft
er v
erif
yin
g
that
Pa
is t
rue
you
mu
st a
ssu
me
that
Pk
is t
rue
for
som
e k
N+.
Th
is a
ssu
mpt
ion
is
calle
d
“In
duct
ion
hyp
oth
esis
”.
A w
ell
- kn
own
illu
stra
tion
use
d to
exp
lain
wh
y th
epr
inci
ple
of
mat
hem
atic
al
indu
ctio
n
wor
ks
is
the
un
endi
ng
dom
ino
line.
Pro
vin
g th
at “
P1
is t
rue”
is
like
knoc
kin
g ov
er t
he f
irst
dom
ina.
Pro
vin
g th
at “
whe
nev
erP
kis
tru
e th
en P
k+
1is
als
o tr
ue”
is
likeh
avin
g th
e do
min
os a
rran
ged
so t
hat
eac
h o
ne
wou
ld k
noc
k do
wn
the
nex
t on
e as
it
fell.
Th
at m
ean
s, y
ou c
ould
kn
ock
them
al
l do
wn
si
mpl
y by
pu
shin
g th
e fi
rst
one.
Mat
hem
atic
al i
ndu
ctio
n w
orks
in
th
e sa
me
way
. S
ince
P1
is t
rue,
th
en P
2m
ust
be
tru
e. S
ince
P2
is t
rue,
P3
mu
st b
e tr
ue.
Sin
ce P
3is
tru
e, h
hen
P4
mu
st a
lso
betr
ue,
an
d so
on
. T
his
su
gges
ts t
hat
you
can
pro
ve a
stat
emen
t P
n
to b
e tr
ue
for
nk
by s
how
ing
that
Pk
impl
ies
Pk
+ 1.
EX
AM
PLE1
1.Fo
r n
= 1
, P1
is t
rue
sin
ce
2.Fo
r n
= k
, ass
um
e th
at
p k:
1 +
2 +
3 +
... +
hol
ds. W
e m
ust
sh
ow t
hat
, for
n=
k+
1,
P k +
1:
1 +
2 +
3 +
... +
(k
+ 1
) =
Now
, add
ing
“k+
1”
to t
he
each
sid
e of
th
e eq
ual
ity
Pk
: 1
+ 2
+ 3
+ ..
. + k
=
we
obta
in
kk
kk
k
kk
kk
2(1)
12
3...
(1)
(1)
23
22
(1)
(2)
.2
kk(
1).
2
kk
(1)
(2)
.2
kk
k(
1).
2
1(1
11
.2
Pro
ve b
y m
ath
emat
ical
indu
ctio
n t
hat
th
e pr
opos
itio
n
is t
rue
for
all n
IN+.
n
nn
Pn
(1)
:12
3...
2
So
luti
on
95
Mat
hem
atic
al In
duct
ion
EX
AM
PLE2
You
may
wor
k ou
t fo
r th
e pr
oof
by y
ours
elf.
Pro
ve b
y th
e m
ath
emat
ical
indu
ctio
n t
hat
a.2
+ 3
+ 6
+ ..
. + 2
n=
n(n
+ 1
)
b.1
+ 3
+ 5
+ ..
. + (
2n–
1) =
n2
are
tru
e fo
r al
l nIN
+.
So
luti
on
EX
AM
PLE3
1.W
hen
n=
1, t
he
form
ula
is v
alid
, bec
ause
2.A
ssu
me
that
hol
ds.
We
mu
st s
how
th
at
Now
, add
ing
(k+
1)2
to e
ach
sid
e of
th
e in
duct
ion
hpp
oth
esis
, we
obta
in
Th
us,
Pk
+ 1
hol
ds w
hen
Pk
is t
rue.
Hen
ce P
nis
tru
e n
IN+.
kk
kk
k
kk
kk
kk
kk
kk
k
kk
k
22
22
2
2
2
(1)
(21)
12
3...
(1)
(1)
6
(1)
(21)
6(1)
6
(1)
[(2
1)6(
1)]
6
(1)
(27
6)6
(1)
(2)
(23)
.6
k
kk
kP
k2
22
21
(1)
(2)
(23)
:12
3...
(1)
.6
k
kk
kP
k2
22
2(
1)(2
1):1
23
...6
P2
1
12
3:1
.6
Use
mat
hem
atic
al in
duct
ion
to
prov
e th
at
for
all n
IN+.
n
nn
nP
n2
22
2(
1)(2
1):1
23
...6
So
luti
on
Th
us,
Pk
+ 1
is t
rue
wh
enev
er P
kis
tru
e. H
ence
.
is t
rue
nIN
+.
n
nn
Pn
(1)
:12
3...
2
96
Appl
ied
Mat
hem
atic
s 11
EX
AM
PLE4
You
may
wor
k ou
t fo
r th
e pr
oof
by y
ours
elf.
Pro
ve b
y m
ath
emat
ical
indu
ctio
n t
hat
for
all n
IN+.
n
nn
Pn
22
33
33
(1)
:12
3...
4
So
luti
on
EX
AM
PLE5
1.Fo
r n
= 1
, is
tru
e.
2.Fo
r n
= k
, ass
um
e
Now
, we
will
sh
ow t
hat
By
hyp
oth
esis
, we
hav
e
By
addi
ng
to e
ach
sid
e of
th
is o
f th
is e
qual
ity,
we
get
Th
at is
, we
hav
e
Th
eref
ore,
Pk+
1is
tru
e. H
ence
, n
IN+:
n
nP
nn
n1
11
1:
...1
22
33
4(
1)1k
kk
kk
k
kk
k
k k
2
2
11
11
21
...1
22
33
4(
1)(
2)(
1)(
2)
(1)
(1)
(2)
(1)
.2
kk
kk
kk
kk
11
11
1...
.1
22
3(
1)(
1)(
2)1
(1)
(2)
kk
1(
1)(
2)
k
kP
kk
k1
11
11
:...
.1
22
3(
1)(
2)2
k
kP
kk
k1
11
11
:...
.1
22
3(
1)(
2)2
k
kP
kk
k1
11
1:
....
12
23
34
(1)
1
P 11
1: 1
21
1
Pro
ve b
y m
ath
emat
ical
indu
ctio
n t
hat
, for
all
posi
tive
nat
ura
l nu
mbe
rs
n
nP
nn
n1
11
1:
....
12
23
34
(1)
1
So
luti
on
97
Mat
hem
atic
al In
duct
ion
EX
AM
PLE6
1.Fo
r n
= 1
, (r
1) is
obv
iosl
y tr
ue.
2.A
ssu
me
that
for
n=
k,
hol
ds. W
e w
ill s
how
th
at P
nis
als
o tr
ue
for
n=
k+
1. N
ow, b
y as
sum
ptio
n
Add
ing
rkto
ea
ch s
ide
of t
he
equ
alit
y ab
ove,
giv
es
Th
at is
, Pn
is t
rue
for
n=
k+
1. H
ence
, n
IN+:
kk
rr
rr
rr
21
11
...,
(1)
1
kk
k
kk
rr
rr
rr
r
rr
rr
21
211
1...
,(
1)1
11
...k r
k
k
rr
r
rr
r
1
1
,(
1)1
1,
(1)
1kk
rr
rr
rr
21
11
...,
(1)
1
kk
k
rP
rr
rr
r2
11
:1...
,(
1)1
rP
r1
1
1:1
,1
Pro
ve b
y m
ath
emat
ical
indu
ctio
n t
hat
nIN
+
nn
n
rP
rr
rr
r2
11
:1...
, (
1).
1
EX
AM
PLE7
Pro
ve b
y th
e m
ath
emat
ical
in
duct
ion
th
at t
he
sum
of
cube
s of
th
ree
con
secu
tive
nat
ura
l
nu
mbe
rs is
div
isib
le b
y 9.
So
luti
on
Let
n, n
+ 1
an
d n
+ 2
be
thre
e co
nse
cuti
ve n
atu
ral n
um
bers
(n
IN+).
Th
e gi
ven
pro
posi
-
tion
in m
ath
emat
ical
for
m is
Pn
: 9
|n3
+ (
n+
1)3
+ (
n+
2)3 .
1.Fo
r n
= 1
,
P1
: 9
| 13
+ 2
3 +
33
. P
1 is
cle
arly
tru
e, s
ince
9
| 36
.
2.A
ssu
me
that
Pk
is t
rue,
i.e.
,
k3+
(k
+ 1
)3+
(k
+ 2
)3
is d
ivis
ible
by
9. W
e w
ill s
how
th
at P
kim
plie
s P
k+
1, i
.e.,
we’
ll pr
ove
that
Pk
+ 1
mu
st a
lso
be t
rue,
9 |
(k+
1)3
+ (
k+
2)3
+ (
k+
3)3
So
luti
on
98
Appl
ied
Mat
hem
atic
s 11
Now
, by
the
assu
mpt
ion
, we
hav
e
k3+
(k
+ 1
)3+
(k
+ 2
)3=
9t
(tIN
+).
Add
ing
–k3
+ (
k+
3)3
to t
he
both
sid
es o
f th
e eq
ual
ity
give
s
k3+
(k
+ 1
)3+
(k
+ 2
)3–
k3+
(k
+ 3
)3=
9t
– k3
+ (
k+
3)3
(k+
1)3
+ (
k+
2)3
+ (
k+
3)3
= 9
t–
k3+
k3
+ 9
k2+
27k
+ 2
7
= 9
t+
9k2
+ 2
7k+
27
= 9
(t+
k2
+ 3
k+
3).
Sin
ce 9
(t+
k2
+ 3
k+
3)
is d
ivis
ible
by
9, w
e m
ay e
asily
con
clu
de t
hat
Pk
+ 1
is
also
tru
e.
Hen
ce, b
y 1°
an
d 2°
Pn
: 9
| n
3+
(n
+ 1
)3+
(n
+ 2
)3
is t
rue
for
all n
IN+.
EX
AM
PLE8
Use
mat
hem
atic
al in
duct
ion
to
prov
e th
at f
or a
ll n
IN+
the
nu
mbe
r “3
2n–
1” is
of
the
form
8k, k
Z.
Th
e pr
opos
itio
n is
Pn
: 8
| 32n
– 1.
1.Fo
r n
= 1
,
P1
: 8
| 32
1–
1 is
cle
arly
tru
e.
2.A
ssu
me
that
Pk
: 8
| 32
k–
1 is
tru
e. N
ow, w
e w
ill p
rove
, for
n=
k+
1,
Pk
+ 1
: 8
| 32
(k+
1)
– 1.
By
hyp
oth
esis
, 32k
– 1
= 8
t(t
Z).
Mu
ltip
lyin
g bo
th s
ides
of
this
equ
alit
y by
32 , w
e ob
tain
32(3
2k–
k) =
32
8t32k
+ 2
– 32
= 7
2t
32(k
+ 1
)–
9 =
72t
,
32(k
+ 1
)–
1 –
8 =
72t
32(k
+ 1
)–
1 =
72
t +
8,
32(k
+ 1
)–
1 =
8(9
t+
1).
It c
an e
asily
be
seen
th
at 3
2(k
+ 1
)–
1 is
als
o of
th
e fo
rm 8
k, k
Z.
i.e.,
Pn
is a
l tr
ue
for
n=
k+
1. T
her
efor
e,
nIN
+.
8 |
32n–
1.
So
luti
on
99
Mat
hem
atic
al In
duct
ion
EX
AM
PLE9
Fin
d th
e tr
uth
set
Na,
aIN
+of
Pn
: 2n
< n
! a
nd
pro
ve it
by
mat
hem
atic
al in
duct
ion
.
1.P
1:
21<
1!
2
< is
not
tru
e.
P2
: 22
< 2
!
4 <
2
is n
ot t
rue.
P3
: 32
< 3
!
8 <
6
is n
ot t
ure
.
P4
: 24
< 4
!
16 <
24
is t
rue.
So
the
tru
th s
et is
N4
{4, 5
, 6 ..
.}
2.A
ssu
min
g th
at P
kis
tru
e, i.
e., P
k: 2
k<
k!,
we
hav
e to
pro
ve t
hat
th
e tr
uth
of
Pk
+ 1
follo
ws
from
th
at o
f P
k. In
oth
er w
ords
, we
will
sh
ow t
hat
Pk
+ 1
: 2k
+ 1
< (
k+
1)!
is t
rue.
It
is c
lear
th
at fo
r k
> 3
, 2 <
k+
1. A
lso,
by
hyp
oth
esis
2k
< k
!. M
ult
iply
ing
thes
e
ineq
ual
itie
s si
de b
y si
de, w
e ge
t
2k· 2
< k
! · (
k+
1)
2k+
1<
(k
+ 1
)!
We
con
clu
de t
hat
Pn
is a
lso
tru
e fo
r n
= k
+ 1
, hen
ce
Pn
: 2n
< n
!C
is t
rue
for
all n
N4.
So
luti
on
EX
AM
PLE10
Pro
ve
“Ber
nou
lli I
neq
ual
ity”
by
mat
hem
atic
al in
duct
ion
.
“Ber
nou
lli I
neq
ual
ity”
nIN
+an
d fo
r 1
+ x
0, (
1 +
x)n
1 +
nx.
1.Fo
r n
= 1
, (1
+ x
)11
+ 1
x
and
also
for
n=
2, (
1 +
x)2
1 +
2
2
Sin
ce
1 +
2x
+ x
21
+ 2
xx2
0
is a
lway
s tr
ue.
2.A
ssu
min
g fo
r n
= k
, Pk
: (1
+ x
)k1
+ k
xh
olds
, we
will
ver
ify
that
Pk
+ 1
: (1
+ x
)k+1
1 +
(k
+ 1
)x
also
hol
ds. B
y h
ypot
hes
is, w
e h
ave
(1 +
x)k
1 +
kx.
Mu
ltip
lyin
g bo
th s
ides
of
this
ineq
ual
ity
by 1
+ x
give
s.
(1 +
x)k
(1 +
x)
(1 +
kx)
(1
+ x
)
(1 +
x)k
+ 1
1 +
x+
kx
+ k
x2
(1 +
x)k
+ 1
1 +
(k
+ 1
)x+
kx2 .
So
luti
on
100
Appl
ied
Mat
hem
atic
s 11
Rec
all b
y th
e tr
ansi
tive
pro
pert
y, a,
b, c
IRif
aan
d b
c, t
he
ac.
By
usi
ng
this
th
is p
rope
rty,
sin
ce
(1 +
x)k
+ 1
1 +
(k
+ 1
)x+
kx2
and,
kx2
0 fo
r k
IN+,
1 +
(k
+ 1
)x+
kx2
1 +
(k
+ 1
)x,
we
can
get
(1 +
x)k
+ 1
1 +
(k
+ 1
)x.
Th
at m
ean
s, P
k+
1is
als
o tr
ue.
Hen
ce, B
ern
oulli
In
equ
alit
y is
tru
e fo
r al
l nIN
+.
BIO
GR
AP
HIC
AL
SK
ET
CH
BE
RN
OU
LLI,
JAK
OP
or J
ACQ
UE
S (
1654
- 1
705)
Sw
iss
mat
hem
atic
ian,
bor
n in
Bas
el.
He
was
the
firs
t of
the
fam
ous
Ber
noul
li fa
mily
to
achi
eve
dist
inct
ion
in m
athe
mat
ics.
He
deve
lope
d th
e ne
wly
inve
nted
cal
culu
s in
to a
pow
erfu
l
mat
hem
atic
al t
ool,
appl
ying
it t
o th
e so
lutio
n of
a v
arie
ty o
f pr
oble
ms.
In
1682
he
open
ed a
sem
inar
y in
Bas
el f
or m
athe
mat
ics
and
expe
rimen
tal p
hysi
cs.
He
was
app
oint
ed p
rofe
ssor
of
mat
hem
atic
s at
the
Uni
vers
ity o
f Bas
el in
168
7. In
his
Ars
con
ject
andi
he
laid
the
foun
datio
ns
of th
e th
eory
of p
roba
bilit
y; B
erno
ulli’
s T
heor
em, g
iven
ther
e, is
of g
reat
impo
rtan
ce w
here
the
num
ber
of “
tria
ls”
is l
arge
. H
e an
d hi
s br
othe
r Jo
hann
wer
e th
e fir
st t
o be
ele
cted
for
eign
asso
ciat
es o
f the
Par
is A
cade
my
of S
cien
ces.
EX
AM
PLE11
Pro
ve t
hat
hep
pro
posi
tion
hol
ds f
or a
ll n
IN+.
n
nP
n2
22
22
:12
3...
(1)
3
1.Fo
r n
= 1
, P1
: (1
– 1
)2 <
is
tru
e.
2.In
duct
ion
hyp
oth
esis
:
Ass
um
e P
kis
tru
e; i.
e.,
Pk
: 12
+ 2
2+
32
+ ..
. + (
k–
1)2
<
hold
s. T
hen,
we
will
pro
ve t
hat
P k+
1is
als
o tr
ue. B
y ad
ding
k2
to b
oth
side
s of
the
ineq
ualit
y
12+
22
+ 3
2+
... +
(k
– 1)
2<
,
we
obta
in
12+
22
+ 3
2+
... +
(k
– 1)
2+
k2
<
+ k
2(I
)k3 3
k3 3
k3 3
3 1 3S
olu
tio
n
101
Mat
hem
atic
al In
duct
ion
EX
AM
PLE12
Use
mat
hem
atic
al in
duct
ion
to
prov
e th
at
nPIN
fIN
ff
nf
nn
n:
,(1
)1,
()
(1)
!.
1.P
1is
tru
e, s
ince
f(2
– 1
) · 2
= 2
!.
2.F
irst
, we
assu
me
for
n=
k,
Pk
: IN
+IN
+, f
(1)
= 1
, f(k
) =
f(k
– 1)
k=
k!.
hol
ds. Y
et, w
e m
ust
sh
ow P
k+1
hol
ds e
ith
er. B
y in
duct
ion
hyp
oth
esis
, we
know
f(k)
= f
(k–
1)k
= k
!
Mu
ltip
lyin
g bo
th s
ides
of
this
equ
alit
y by
k+
1 g
ives
f(k
– 1)
· k
(k+
1)
= k
! · (
k+
1)
= f
(k)
(k+
1)
= k
+ 1
)!.
Th
at is
Pk+
1:
IN+
IN+, f
(1)
= 1
, f(k
+ 1
) =
f(k
) (k
+ 1
) =
(k
+ 1
)!
Hen
ce, P
nis
tru
e n
IN+.
f
f
So
luti
on
Now
, it
is c
lear
th
at f
or e
very
kIN
+
k3+
3k2
< k
3+
3k2
+ 3
k+
1
k3+
3k2
< (
k+
1)3 ,
By
usi
ng
the
tran
siti
ve p
rope
rty,
from
th
e in
equ
alit
ies
I an
d II
, we
get
Th
us,
Pk
+ 1
is a
lso
tru
e. H
ence
, we
can
con
clu
de t
hat
nn
32
22
21
23
...(
1).
3
kk
22
22
2(
1)1
23
....
3
kk
k
kk
k
32
3
33
2
3(
1),
33
(1)
33
(II)
102
Appl
ied
Mat
hem
atic
s 11
EX
AM
PLE13
For
n3,
Pn
: 2n
> 2
n+
1. P
rove
th
at P
nis
tru
e fo
r al
l nIN
+.
1.Fo
r n
= 3
, P3
: 23
> 2
· 3
+ 1
is c
lear
ly t
rue.
2.A
ssu
min
g
Pk
: 2k
> 2
k+
1
is t
rue,
we
will
ver
ify
that
Pk
+ 1
is a
lso
tru
e. N
ow,
for
k3
it i
s cl
ear
that
2k
> 2
an
d by
hyp
oth
esis
, we
hav
e
2k>
2k
+ 1
.
Add
ing
thes
e tw
o in
equ
alit
ies
side
by
side
yie
lds.
2k+
2k
> 2
k+
3
2k+
1>
2(k
+ 1
) +
1.
Th
eref
ore,
Pk+
1is
als
o tr
ue.
Hen
ce,
n3,
2n
> 2
n+
1.
So
luti
on
EX
AM
PLE14
Pro
ve “
De
Moi
vre’
s T
heo
rem
” by
mat
hem
atic
al in
duct
ion
.
1.Fo
r n
= 1
, [(r
cos
+ i
sin
)
]1=
r(c
os (
1)
+ i
sin
(1
) is
tru
e.
2.A
ssu
min
g th
at t
he
form
ula
[r(c
os
+ i
sin
)
]k=
rk
(cos
k+
isi
n k
)
is t
rue,
we
mu
st s
how
th
at t
he
form
ula
[r(c
os
+ i
sin
)
]k+
1=
rk
+ 1
[cos
(k
+ 1
)+
isi
n (
k+
1)
)
also
hol
ds. N
ow, b
y in
duct
ion
hyp
oth
esis
, we
hav
e
[r(c
os
+ i
sin
)k
= r
k(c
os k
+ i
sin
k).
If w
e m
ult
iply
eac
h s
ide
of t
his
equ
alit
y by
r(c
os+
isi
n
), w
e ob
tain
[r(c
os
+ i
sin
)]
k[r
(cos
+
isi
n
)]=
rk
(cos
k+
isi
n k
) [r
(cos
+
isi
n
)],
[r(c
os
+ i
sin
)]
k+
1=
rk
+ 1
[co
s (k
+
) +
isi
n (
k+
)]
= r
k+
1[c
os (
k+
1)
+ i
sin
(k
+ 1
) ].
Hen
ce, f
or a
ll po
siti
ve in
tege
r n
,
[r(c
os
+ i
sin
)]
n=
rn
(cos
n+
isi
n n
).
So
luti
on
Th
eo
rem
If z
= r
(cos
+
isi
n
) is
a c
ompl
ex n
um
ber
and
n is
a p
osit
ive
inte
ger,
then
zn=
[r
(cos
+
isi
n
) ]n
= r
n(c
os n
+ i
sin
n).
De
Moi
vre’
s T
heo
rem
103
Mat
hem
atic
al In
duct
ion
EX
AM
PLE15
Pro
ve b
y m
ath
emat
ical
indu
ctio
n t
hat
n
IN+, n
! n
n–
1 .
EX
AM
PLE16
Pro
ve “
Bin
omia
l Th
eore
m”
by m
ath
emat
ical
indu
ctio
n.
1.Fo
r n
= 1
, P1
: 1!
11
– 1 , f
or n
= 2
, P2
: 2!
n
2 –
1an
d fo
r n
= 3
, P3
: 3!
33
– 1 . T
hu
s P
1, P
2
and
P3
are
all t
rue.
2.A
ssu
me
that
, for
n=
k,
Pk
: k!
kk
– 1
is t
rue.
Now
, we
will
sh
ow t
hat
Pk
+ 1
: (k
+ 1
)!
(k+
1)k
also
hol
ds. W
e kn
ow, b
y in
duct
ion
hyp
oth
esis
, th
at k
! kk
– 1 . M
ult
iply
ing
both
sid
es o
f th
is
ineq
ual
ity
by k
+ 1
yie
lds.
k! ·
(k+
1)
kk–
1· (
k+
1)
= (
k+
1)!
kk
+ k
k–
1(I
).
By
usi
ng
bin
omia
l exp
ansi
on
we
can
obt
ain
th
e fo
llow
ing
ineq
ual
ity
(k+
1)!
(k
+ 1
)k .
Th
us,
Pn
is t
rue
for
n=
k+
1. H
ence
, we
can
con
clu
de t
hat
nIN
+, n
! n
n–
1 .
kk
kk
kk
kk
kk
k1
2(
1)(
)(
)(
)...
1,0
12
So
luti
on
Th
eo
rem
If a
and
bar
e re
al n
um
bers
an
d n
is a
pos
itiv
e in
tege
r, th
en
wh
ere
is t
he
bin
omin
al c
oeff
icie
nt
nn k
kn
k!
()
.!(
)!n k(
)
nn
nn
nn
nn
nn
ab
ab
ab
abb
nn
12
21
()
()
()
()
...(
)(
)0
12
1
Bin
omia
l Th
eore
m
1.Fo
r n
= 1
, P1
: (a
+ b
)1=
S
ince
bot
h c
oeff
icie
nts
equ
al t
o 1,
th
e eq
uat
ion
redu
ces
to (
a+
b)1
= a
+ b
.
2.N
ext,
su
ppos
e th
at t
he
form
ula
hol
ds f
or n
= k
, i.e
.,
kk
kk
kk
kk
kk
Pa
ba
ab
abb
kk
11
:()
()
()
...(
)(
).
01
1
ab
11
11
()
()
.0
1S
olu
tio
n
104
Appl
ied
Mat
hem
atic
s 11
Now
we
use
th
e pr
oper
ty t
o sh
ow t
hat
is t
rue.
To
do
this
, mu
ltip
ly b
oth
sid
es o
f th
e fo
rmu
la f
or n
= k
by a
+ b
to o
btai
n
Add
ing
like
ter
ms,
su
ch a
s an
d ak
b ,
usi
ng
the
prop
erty
give
s
Hen
ce, P
nis
tru
e n
N+.
kk
kk
kk
ka
ba
ab
bk
11
11
11
()
)(
)(
)...
()
.0
11
kk
kr
rr
1(
)(
)(
)1
kk
ab
()
,0
kk
ab
k()
kk
kk
k
kk
kk
kk
kk
ab
aa
ba
bab
k
kk
kk
ab
ab
abb
kk
11
12
12
1
()
()
()
()
...(
)0
12
()
()
...(
)(
).
01
1
kk
kk
k
kk
kP
ab
aa
bb
k1
11
1
11
1:(
)(
)(
)...
()
.0
11
EX
AM
PLE17
Pro
ve t
hat
n
N+.
cos
· cos
2· c
os 4
· ...
· cos
2=
n
n
1
1
sin
2.
2si
n
1.Fo
r n
= 1
, P1
: co
s · c
os 2
a=
cos
· cos
2=
4 · s
in
· cos
· c
os 2
= s
in 4
,
2 · 2
· si
n
· cos
· c
os 2
= s
in 4
,
2 · s
in 2
· cos
2=
sin
4.
So,
P1
is t
rue.si
n2
sin
4,
4si
n
11
11
sin
2,
2si
nS
olu
tio
n
105
Mat
hem
atic
al In
duct
ion
2.A
ssu
me
that
Pk
: co
s co
s 2
... c
os 2
k=
is tr
ue, w
e w
ill p
rove
Pk
impl
ies
P k+
1. M
ultip
lyin
g bo
th s
ides
of t
he e
quat
ion
P kby
cos
2k
+ 1
yiel
ds.
Hen
ce, P
k+
1h
olds
; th
at is
, Pn
is t
rue
nIN
+.
kk
kk
k
k
k
k
k
11
11
1
1 (1)
1
(1)
1
sin
2co
s2co
sco
s2...
cos2
cos2
,2
sin
1si
n2
22
,2
sin
sin
2.
2si
n
k
k
1
1
sin
22
sin
EX
AM
PLE18
Pro
ve t
hat
for
all
nIN
+
n
n
times
12
22
...2
22
cos
.2
1.Fo
r n
= 1
, is
tru
e.
2.A
ssu
min
g fo
r n
= k
,
is t
rue,
we
will
sh
ow t
hat
Pk
+ 1
also
hol
ds. N
ow, b
y ad
din
g “2
” to
bot
h s
ides
of
the
equ
ali-
ty g
iven
abo
ve, w
e ge
t
Taki
ng
the
squ
are
root
of
the
both
sid
es y
ield
s
Sin
ce c
os 2
= 1
k1
2(c
os2
cos
).2
k
k
1tim
es
12
22
...2
2(1
cos
),2
k
k
times
12
22
2...
22
22
cos
.2
k
kk
Ptim
es
1:
22
2...
22
2co
s2
P 11
1
2:
22
cos
cos
24
2S
olu
tio
n
106
Appl
ied
Mat
hem
atic
s 11
Rec
all f
rom
“A
lgeb
ra 3
” th
at
By
usi
ng
this
con
vers
ion
for
mu
la, w
e ob
tain
Hen
ce,
is t
rue
nIN
+.
n
nn
Ptim
es
1:
22
2...
22
2co
s2
kk
k
kk
kk
kk
k
11
1
22
22
22
2
(1)
122
22
22
...2
22
cos(
)cos
()
22
22
cos(
)cos
()
22
2(
cos
)(co
s)
22
2co
s2
cos
22
2co
s.
2
xy
xy
xy
cos
cos
2co
s()
cos(
).2
2
EX
AM
PLE19
Pro
ve b
y us
ing
the
Gau
sian
met
hod
that
n li
nes
, no
two
are
para
llel,
inte
rsec
t at
mos
t C
(n, 2
)po
ints
in a
pla
ne.
Two
lines
d1
an d
2w
hic
h a
re n
ot p
aral
lel t
o ea
ch o
ther
in t
erse
ct a
t
mos
t on
e po
int,
say
A.
Th
e lin
e d 3
wh
ich
is
not
par
alle
l to
an
y on
e of
th
e lin
es d
1an
d d 2
inte
rsec
ts t
hes
e lin
es a
t m
ost
two
poin
ts, s
ay B
and
C.
Th
e lin
e d 4
wh
ich
is
not
par
alle
l to
an
y of
th
e lin
es d
1, d 2
and
d 3in
ters
ects
th
ese
lines
at
mos
t th
ree
poin
ts, s
ay D
, Ean
d F.
Con
tinui
ng i
n th
is w
ay,
for
ne
IN,
the
line
dn i
nter
sect
s th
e lin
es
d 1, d
2, ...
dn
– 1
at m
ost
(n–
1)po
ints
.
If w
e ad
d th
e n
um
ber
of a
ll th
e po
ints
for
med
by
the
inte
rsec
tion
of
the
lines
giv
en a
bove
, th
en w
e ob
tain
inte
rsec
tion
poi
nts
in a
pla
ne.
nn
nC
n(
1)1
23
...(
1)(
, 2)
2
So
luti
on
A
d 1
d 2
A
d 1
d 2
d 3
CB
A
d 1
d 2
d 3
C
B
FED
d 4
107
Mat
hem
atic
al In
duct
ion
EX
AM
PLE20
Pro
ve b
y m
ath
emat
ical
in
duct
ion
th
at b
y jo
inin
g n
poin
ts,
no
thre
e ar
e co
llin
ear,
we
can
fro
m C
(n,
2)
diff
eren
t lin
es.
1.Fo
r n
= 2
, C(2
, 2)
= 1
is t
rue,
sin
ce t
her
e is
on
ly o
ne
line
pass
ing
thro
ugh
tw
o po
ints
.
2.A
ssu
me
that
n=
kpo
ints
, no
thre
e ar
e co
llin
ear,
we
can
form
C(k
, 2)
diff
eren
t lin
es. N
ow
we
will
sh
ow t
hat
for
n=
k+
1 p
oin
ts C
(k+
1, 2
) di
ffer
ent
lines
can
be
form
ed.
Add
ing
one
mor
e po
int
form
s k
extr
a lin
es. S
o, w
e w
ill h
ave
diff
eren
t lin
es. H
ence
, we
can
for
m
diff
eren
t lin
es b
y jo
inin
g n
poin
ts, n
o th
ree
are
colli
nea
r.
nn
Cn
(1)
(,
2)2k
kk
kk
kk
Ck
kk
Ck
(1)
(1)
2(
1)(
2)(
1,2)
22
2
So
luti
on
1.Fo
r n
= 3
, C
(3,
3) =
1 i
s tr
ue
sin
ce o
nly
on
e tr
ian
gle
can
be
draw
n u
sin
g th
ree
poin
ts
2.A
ssu
me
that
C(k
, 3)
tria
ngl
es c
an b
e fo
rmed
usi
ng
k po
ints
lie
on a
cir
cle.
Yet
we
will
sh
ow,
C(k
+ 1
, 3)
tria
ngle
s ca
n be
dra
wn
usin
g k
+ 1
poi
nts
lyin
g on
a c
ircl
e. B
y pr
evio
us p
robl
em, w
e kn
ow th
at
C(n
, 2)
diff
eren
t lin
es c
an b
e fo
rmed
by
join
ing
npo
ints
, n
o th
ree
are
colli
nea
r. S
o ad
din
g on
e
mor
e po
int
form
s C
(k, 2
) ex
tra
tria
ngl
es. T
hat
is
C(k
, 3)
+ C
(k, 2
) =
C(k
+ 1
, 3)
tria
ngl
es c
an b
e fo
rmed
usi
ng
k+
1 p
oin
ts ly
ing
on a
cir
cle.
Hen
ce,
tria
ngl
es c
an b
e dr
awn
by
usi
ng
npo
ints
lie
on a
cir
cle.
In
vest
igat
e th
e F
igu
re 1
.2 d
raw
n
for
n=
9 p
oin
ts.n
nn
Cn
(1)
(2
(,
3)6
So
luti
on
EX
AM
PLE21
Pro
ve b
y m
ath
emat
ical
in
duct
ion
th
at C
(n,
3) t
rian
gles
can
be
draw
n b
y u
sin
g th
e ve
rtic
es c
hos
en f
rom
n p
oin
ts li
e on
a c
ircl
e.
I
II
III
IV
V
VI
VII
VIII
IX
108
Appl
ied
Mat
hem
atic
s 11
EX
ER
CIS
ES
3.1
1.U
se m
ath
emat
ical
in
duct
ion
to
prov
e th
e fo
llow
-
ing
equ
alit
ies
for
ever
y po
siti
ve in
tege
r n
.
a. b. c. d.1
+ 7
+ 1
3 +
19
+ ..
. + 6
n–
5 =
n(3
n–
2)
e. f.13
+ 3
3+
53
+ 7
3+
...+
(2n
– 1)
3=
n2 (2
n2
– 1)
g. h.21
2 +
22
3 +
23
4 +
...+
2n (n
+ 1
) =
n2n+
1
i. j. k. l. m.
n. o.n
nn
n1
11
1...
13
35
57
(21)
(21)
21
nn
23
11
11
1...
12
22
22
nn
nn
11
31
25
13...
(23
)2
12
nn
21
31
13
3...
32
kn
nn
k
nn
nn
kk
12.
.....
(1)
...(
1)
(1)
(2)
...(
)(
1)
nn
n
nn
nn
12
32
34
...(
1)(
2)
(1)
(2)
(3)
4
nn
nn
n(
1)(
2)1
22
3...
(1)
3
nn
nn
n2
22
21
21
(1)
12
34
...(
1)(
1)2
nn
n2
22
22
2(4
1)1
35
7...
(21)
3
nn
n(3
1)1
47
10...
32
2
nn
n3
(1)
36
9...
32
nn
n(3
1)2
58
11...
31
2
3.U
se m
ath
emat
ical
in
duct
ion
to
prov
e th
e fo
llow
-
ing
ineq
ual
itie
s fo
r th
e in
dica
ted
posi
tive
in
tege
r
valu
es o
f n
.
a.2n
1 +
n,
n+
b.2n
> 2
n+
1,
n+
c.3n
n2n
, n
+
d.4n
> n
2 , n
+
e.5n
1 +
4n
, n
+
f.(2
n)!
2n
(n!)
2 , n
+
g.n
! n
n–
1 , n
+
2.F
ind
the
tru
th s
et o
f th
e fo
llow
ing
prop
osit
ion
s,
and
use
mat
hem
atic
al in
duct
ion
to
prov
e th
em.
a.2n
n2
b.3n
1 +
4n
c.n
3>
n2
+ 3
d.n
n11
!2
o. p. r. s. t.1
1! +
2
2! +
3
3! +
nn!
= (
n+
1)!
– 1
u.n n
n1
23
11
01
2!3!
4!!
!nn
nn
nn
22
22
12
3(
1)...
13
35
57
(21)
(21)
2(2
1)
nn
nn
nn
nn
56
4...
12
32
34
()(
2
(37)
2(1)
(2)
nn
nn
11
1...
14
47
(32)
(31)
31
nn
nn
11
11
...1
33
55
7(2
1)(2
1)2
1
109
Mat
hem
atic
al In
duct
ion
10.
Pro
ve t
hat
n
+an
d a,
x, y
+
f(x
y) =
f(x
) f(
y)
f(an
) =
[ f
(a)]
n.
12.
Pro
ve t
hat
a p
olyg
on w
ith
n-s
ides
(n
> 3
) h
as
diag
onal
s.(
3)2
nn
11.
Pro
ve t
hat
n
+an
d a,
x, y
+
f(x
y) =
f(x
) +
f(y
) f(
an)
= n
f(a)
.
13.
If x
1>
0, x
2>
0, .
.., x
n>
0, t
hen
prov
e th
at
n+
In (
x 1x 2
x 3...
xn
) =
In
x1
+ I
n x
2+
... +
In
xn
.
5.P
rove
th
at
n+
nn
1co
sco
s()
22
sin
sin
2...
sin
()
.2
sin
2
6.P
rove
th
at
n+
nn
n
(1)
sin
cos(
)2
2co
sco
s2...
cos(
).
sin
2
7.P
rove
th
at
n+
sin
2co
sco
s3...
cos(
21)
.2
sinn
n
8.P
rove
th
at
n+
23
1(
1)lo
g()
2lo
g()
...lo
g()
log(
).1
2!
nn
nn
nn
9.P
rove
th
at
n+
()
...[
(1)
][2
(1)
].2n
xx
yx
ny
xn
y
4.U
se m
ath
emat
ical
in
duct
ion
to
prov
e th
e fo
llow
-
ing
expr
essi
ons
for
ever
y po
siti
ve in
eger
n.
a.5
| 6n
– 1
b.6
| 7n
– 1
c.2
| 22n
+ 2
d.3
| 22n
+ 2
e.3
| n
3–
nf.
7 |
132n
+ 6
g.11
| 2
4n–
5nh.
11 |
12n
+ 1
0
i.3
| 4n
– 1
j.5
| 5n
– 3n
k.(a
+ b
) |
a2n–
b2n
l.27
| (
25n+
1+
5n
+ 2)
m. 6
| (
5n3
+ 1
3n–
24)
n.5
| 17
4n+
1+
3 ·
92n
o.7
| 32n
+ 2
– 2n
+ 1
p.13
3 |
11n
+ 2
+ 1
22n+
1
r.(x
+ y
) |
x2n–
1 + y
2n–
1
s.6
| n
(2n
+ 1
) (7
n+
1)
h.Fo
r x 1
, x2,
x 3...
xn
and
n+
|x1
+ x
2+
... +
xn
| |x
1| +
|x 2
| +
... +
|x n
|
i.2!
4!
6!
... (
2n)!
[(
n+
1)!
]n,
n+
j. k.n
n1
11
1...
1,
39
273n
n1
11
1...
1,
24
82
110
Appl
ied
Mat
hem
atic
s 11
Add
itio
n i
s as
fu
nda
men
tal
in a
dvan
ced
mat
hem
atic
s as
it
is i
n a
rith
met
ic.
Th
ere
is a
sto
ry
ofte
n t
old
abou
t th
e G
erm
an m
ath
emat
icia
n K
arl
Frie
dric
h G
auss
. W
hen
he
was
in
th
ird
grad
e, h
is c
lass
mis
beh
aved
an
d th
e te
ach
er g
ave
the
follo
win
g pr
oble
m a
s pu
nis
hm
ent.
“Add
th
e in
tege
rs f
rom
1 t
o 10
0.”
It i
s sa
id t
hat
Gau
ss s
olve
d th
e pr
oble
m i
n a
lmos
t n
o ti
me
at a
ll. H
is m
eth
od w
as o
f co
urs
e
diff
eren
t. I
n t
his
sec
tion
we
will
stu
dy s
um
s of
ter
ms
of v
ario
us
sequ
ence
s, l
earn
a s
peci
al
not
atio
n f
or s
um
s an
d al
so f
or p
rodu
cts
of t
hei
r te
rms.
A.
SU
MM
ATIO
N N
OTA
TIO
N
A c
onve
nie
nt
not
atio
n f
or t
he
sum
of
the
term
s of
a g
iven
fu
nct
ion
def
ined
as
f:
Z, f
(k)
=a k
is s
um
mat
ion
not
atio
n o
r si
gma
not
atio
n, w
hic
h in
volv
es t
he
Gre
ek c
apit
al le
tter
sig
ma
().
Defi
nit
ion
Let
: Z
, f(k
) =
ak
and
r, n
Zpr
ovid
ed t
hat
rn
, th
en
is t
he
sum
of
(n–
r+
1)
term
s of
th
e fu
nct
ion
ffr
om r
to n
wh
ere
kis
th
e in
dex
of s
um
ma-
tion
, ris
th
e lo
wer
bou
nd
and
nis
th
e u
pper
bou
nd.
12
...n
kr
rr
nk
r
aa
aa
a
Not
e th
at,
can
als
o be
den
oted
by
and
is r
ead
as
“th
e su
m o
f a k
from
k=
rto
k=
n”.
Stu
dy t
he
exam
ples
giv
en b
elow
.18
56
718
57
32
17
3
22
22
1 110
3 1
...,
...,
12
32...
,
13
5...
21
(21)
,
(2)
68
1012
1416
1820
,
10
12
3...
.
kk
kk
n k n k
kn
kaa
aa
a
aa
aa
a
nk
nk
k
kn
[,
]
or
n
kk
rk
nk
rn
aa
n
kk
r
a
111
Mat
hem
atic
al In
duct
ion
1.
Su
mm
ati
on
Fo
rm
ula
s
a.b.
c.1
(2)
(1)
n k
kn
n2
1
(21)
n k
kn
1
(1)
2
n k
nn
k
FO
RM
UL
A
2
1
(1)
(21)
.6
n k
nn
nk
FO
RM
UL
A
We
will
pro
ve t
he
firs
t an
d th
e th
ird
form
ula
, you
may
wor
k ou
t fo
r th
e pr
oof
of s
econ
d ca
se
by y
ours
elf.
Pro
of
1.
Add
ing
n+
(n
– 1)
+ ..
. + 1
to
th
e ex
pres
sion
giv
en a
bove
, wh
ich
als
o eq
ual
to
P, g
ives
3.
By
subs
titu
tin
g th
e “F
orm
ula
1”
into
th
e ab
ove
expr
essi
on, w
e ge
t
1
(1)
22
(1)
2
n k
kk
Tk
kk
1
Let
2
24
6...
22(
12
3...
n k
Tk
nn
times
2(
1)(
1)...
(1)
(1)
2(
1),
(1)
.2
n
Pn
nn
n
Pn
n
nn
P
1
Let
1
2...
(1)
.n k
Pk
nn
Pro
of
In o
rder
to
deri
ve t
his
for
mu
la w
e w
ill u
se t
he
expa
nsi
on
(x+
y)3
= x
3+
3x2 y
+ 3
xy2
+ y
3
It is
cle
ar t
hat
(1 +
3)3
= 1
3+
3 ·
12· 1
+ 3
· 1
· 12
+ 1
3 ,
(2 +
1)3
= 2
3+
3 ·
22· 1
+ 3
· 2
· 12
+ 1
3 ,
(3 +
1)3
= 3
3+
3 ·
32· 1
+ 3
· 3
· 12
+ 1
3 ,
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
......
(n+
1)3
= n
3+
3 ·
n2
· 1 +
3 ·
n· 1
2+
13 .
112
Appl
ied
Mat
hem
atic
s 11
Add
ing
all o
f th
ese
equ
alit
ies
side
by
side
yie
lds
(n+
1)3
= 1
3+
3 ·
(12
+ 2
2+
32
+ ..
. + n
2 ) +
3 ·
(1 +
2 +
3 +
... +
n)
+ n
· 13
....
(I)
Sin
ce
We
can
wri
te t
he
expr
essi
on (
I) a
s
If w
e re
arra
nge
th
is e
qual
ity,
we
will
obt
ain
Hen
ce,
2
1
(1)
(21)
.6
n k
nn
nk
23
22
32
1
2
62
64
33
23
(23
1)(
1)(2
1).
n k
kn
nn
nn
nn
n
nn
nn
nn
32
2
11
32
2
1
33
11
33
,
(1)
32
33
.2
nn
kk
n k
nn
nk
kn
nn
nn
nk
22
22
2
11
12
3...
an
d
12
3...
,n
n
kk
kn
kn
32
1
(1)
()
.2
n k
nn
k
FO
RM
UL
A
Pro
of
We
can
pro
ve F
orm
ula
1.3
by
calc
ula
tin
g th
e ar
ea o
f
the
squ
are
wit
h s
idel
ngt
h 1
+ 2
+ .
.. +
nu
nit
s. I
n
Fig
ure
1.1
, th
e le
tter
s A
1, A
2, A
3, ...
, A
nde
not
e th
e
area
s pa
inte
d by
th
e sa
me
colo
r.
A(K
LM
N)
= A
1+
A2
+ A
3+
... +
An
= (
1 +
2 +
3 +
... +
n)2
32
1
33
33
2
18
27...
() (
1)1
23
...(
).
2
n k
nk n
nn
NM L
1
n 23 1 K2
3n
A1
A2
A3
An
113
Mat
hem
atic
al In
duct
ion
1
1
1,
11
nn
k
k
rr
rr
FO
RM
UL
A
1
1(
1)1
n k
nk
kn
FO
RM
UL
A
Pro
of
... (
I)
Mu
ltip
lyin
g bo
th s
ides
of
this
equ
alit
y by
r, w
e ob
tain
r·T
= r
+ r
2+
r3
+ ..
. + r
n=
1
... (
II)
Now
, if
we
subt
ract
th
e eq
ual
ity
(II)
from
th
e eq
ual
ity
(I)
side
by
side
, we
will
get
T–
rT=
1 –
rn
T(1
– r
) =
1 –
rn
1,
(1)
.1
n rr
r
12
1
1
Let
1
...n
kn
k
Tr
rr
r
Pro
of
Sin
ce
we
obta
in
11
11
1(
)(
1)1
11
11
11
11
...1
22
33
41
11
.1
1
nn
kk
kk
kk
nn
nn
n
11
1,
(1)
1k
kk
k
1
11
11
1...
.(
1)1
22
33
4(
1)
n kk
kn
n
Pro
of
It is
obv
iou
s th
at
and
1(
1) ti
mes...
(1)
.n k
nr
cc
cc
cn
rc
1 ti
mes
...n k
n
cc
cc
cn
c
2.
Pro
perti
es o
f S
um
mati
on
No
tati
on
Pro
perti
es 1
Let
can
d r
,
1
an
d
(1)
.n
n
kk
r
cn
cc
nr
c
114
Appl
ied
Mat
hem
atic
s 11
EX
AM
PLE22
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
c.d.
13
7
1 ()
8n
12
4
5t
50
1
(2)
p
17
1
3k
Her
e, w
e w
ill s
olve
on
ly “
a” a
nd
“c”.
Oth
ers
are
left
as
an e
xerc
ise
to t
he
stu
den
t.
a.c.
12
4
55
(12
41)
45t
17
1
33
1751
k
So
luti
on
EX
AM
PLE23
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
c.6
3
1
(2
)p
p10
2
1
3k
k11
1
2k
k
a.
b.an
d c
.ar
e le
ft a
s an
exe
rcis
e fo
r th
e st
ude
nt.
1111
11
1112
22
213
2.2
kk
kk
So
luti
on
Pro
of
It is
obv
iou
s th
at
12
31
12
3
1
...
(...
)
.
n
kn
k
n
n
kk
ca
caca
caca
ca
aa
a
ca
Pro
perti
es 2
For
c,
11
.n
n
kk
kk
ca
ca
Pro
of
It is
obv
iou
s th
at
12
11
1
1 1
......
,
.
n
kp
pp
nk
pn
kk
kk
p
aa
aa
aa
a
aa
Pro
perti
es 3
For
1 <
p<
nan
d p
, 1
11
.p
nn
kk
kk
kk
p
aa
a
115
Mat
hem
atic
al In
duct
ion
Pro
of
It is
obv
iou
s th
at
11
22
1
12
12
11
()
()
()
...(
)
(...
)(
...)
.
n
kk
nn
k
nn
nn
kk
kk
ab
ab
ab
ab
aa
ab
bb
ab
Pro
perti
es 4
not
atio
n h
as d
istr
ibu
tive
pro
pert
y ov
er a
ddit
ion
an
d su
btra
ctio
n, i
.e.,
11
1
()
.n
nn
kk
kk
kk
k
ab
ab
EX
AM
PLE24
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
201
10
2k
k
47
24
1(
1)k
kk
a. b.20
209
11
1
1010
1
209
209
22
2
12
12
22
.1
21
2
kk
k
kk
k
4747
23
241
1
11
1(
1)(
1)(
1)
4723
4824
4746
1.
4848
48
kk
kk
kk
kk
kS
olu
tio
n
EX
AM
PLE25
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
62
1
(21)
k
k5
32
1
(2
)k
kk
a. b.It
left
as
an e
xerc
ise
for
the
stu
den
t.
55
53
23
2
11
1
55
32
2
11
(2
)2
56
56
112
()
233
5.2
6
kk
k
kk
kk
kk
kk
So
luti
on
116
Appl
ied
Mat
hem
atic
s 11
Note
not
atio
n d
oes
not
hav
e di
stri
buti
ve p
rope
rty
over
mu
ltip
licat
ion
an
d di
visi
on, i
.e.,
11
11
11
()
an
d
(:
):
.n
nn
nn
n
kk
kk
kk
kk
kk
kk
kk
ab
ab
ab
ab
Pro
perti
es 5
Let
r, p
then
You
may
wor
k ou
t fo
r th
e pr
oof
as a
n e
xerc
ise..
nn
rn
r
kk
rk
rk
pk
pr
kp
r
aa
a
EX
AM
PLE26
Eva
luat
e.
a.b.
c.3
2
0
(32)
k
k8
1(24)
k
k7
2
3
(1)
k
k
a. b. c.It
is le
ft a
s an
exe
rcis
e fo
r th
e st
ude
nt.
88
210
11
21
10
1
(24)
2(2)
4(2
44)
1011
22
110.
2
kk
k
k
kk
k
k
77
25
22
2
33
21
52
1
(1)
(2
1)(
1)
56
115
6(
21)
25
90.
62
kk
k
k
kk
k
kk
So
luti
on
Pro
of
12
31
11 11
121
2122
21
2
12
11
1
12
11
1
(...
)
(...
)(
...)
...(
...)
...
(...
)(
).
nm
n
krk
kk
kmk
rk
mm
nn
nm
mm
m
rr
nrr
rr
mm
n
rr
nrkr
rr
k
aa
aa
a
aa
aa
aa
aa
a
aa
a
aa
aa
Pro
perti
es 6
Com
mu
tati
vity
of
11
11
()
nm
mn
krkr
kr
rk
aa
117
Mat
hem
atic
al In
duct
ion
EX
AM
PLE27
Eva
luat
e th
e fo
llow
ing
expr
essi
ons.
a.b.
43
10
()
ab
ab
53
2
11
(2
)k
r
kr
a.
Als
o,
b.T
he
solu
tion
is le
ft a
s an
exe
rcis
e fo
r th
e st
ude
nt.
35
32
11
1
56
11(
2)
(5
2)
6
34
553
1016
560
105.
2
rk
r
kr
r
53
52
2
11
1
55
52
2
11
1
34
(2
)(3
2)
2
(312
)3
12
56
113
125
165
6010
5.6
kr
k kk
k
kr
k kk
So
luti
on
EX
AM
PLE28
Fin
d go
f (2
) if
2
13
11
10
1(
)(2
) a
nd
()
().
14
xx
xx
km
pr
fx
km
gx
pr
Sin
ce g
o f
(2)
= g
( f(
2)),
we
nee
d to
eva
luat
e g(
4).
57
10
58
11
5
1
5
1
(4)
()
(1)
89
(88)
2
56
(828
)8
285
2
260.
pr
pr
p p
gp
r
pr
p p
2
11
22
11
24
11
1(2
)(2
)14 1
45
1(8
)(8
10)
142
14
11
[8(
2)10
](8
6)14
14
14
51
(86
4)56
4.14
214
xx
km
kk
kk
fk
m
kk
kk
So
luti
on
118
Appl
ied
Mat
hem
atic
s 11
B.
MU
LTIP
LIC
ATIN
NO
TA
TIO
N
In t
he
prec
edin
g se
ctio
n, w
e h
ave
disc
uss
ed t
he
sum
of
the
term
s of
a f
un
ctio
n g
iven
as:
f:
Z, f
(k)
= a
k
Rem
embe
r th
at, w
e u
sed
(sig
ma
not
atio
n)
in o
rder
to
den
ote
the
sum
of t
he
term
s of
th
e
fun
ctio
n f.
Now
, we
will
intr
odu
ce a
new
not
atio
n,
(pi)
, use
d fo
r re
pres
enti
ng
the
prod
uct
of t
he
term
s of
th
e ab
ove
fun
ctio
n.
Defi
nit
ion
Let
f:
Z, f
(k)
= a
kan
d r,
nZ
prov
ided
th
at r
n, t
hen
is t
he
prod
uct
of
(n–
r+
1)
term
s of
th
e fu
nct
ion
ffr
om r
to n
wh
ere
kis
th
e in
dex
of
mu
ltip
licat
ion
, r is
th
e lo
wer
bou
nd
and
n is
th
e u
pper
bou
nd.
12
...n
kr
rn
kr
aa
aa
can
als
o be
den
oted
by
and
is r
ead
as “
prod
uct
of
a k’s
fro
m k
= r
to k
= n
”
Stu
dy t
he
follo
win
g ex
ampl
es.
40
34
540
3
0
98
70
9
1
203
33
33
1 1
100
12
310
0
1
...,
...,
12
3...
,
12
34
..., 1
12
3...
,1
23
41
33
33
...3
.
kk
kk
n k
k n k
k
kaa
aa
a
aa
aa
a
nk
nk
n
kn
kn
[,
]
or
n
kk
rk
nk
rn
aa
n
kk
r
a
1.
Mu
ltip
licati
on
Fo
rm
ula
s
a.b.
!(
1)!
n
kp
nk
p1
!n k
kn
FO
RM
UL
A
119
Mat
hem
atic
al In
duct
ion
2.
Pro
perti
es o
f M
ult
ipli
cati
on
No
tati
on
Pro
of
1.B
y de
fin
itio
n,
2.O
bvio
usl
y, (
1)(
2)..
12
3...
(1)
(1)
...1
23
...(
1)
!.
(1)
!
n
kp
kp
pp
n
pp
pn
p
np
1
12
3...
,
!
n k
kn
n
log
(1)
log
(1)
.n
kp
kp
kn
FO
RM
UL
A
Pro
of
Her
e, w
e sh
all u
se t
he
prop
erti
es o
f lo
gari
thm
.
1lo
g(
1)lo
g(
1)lo
g(
2)...
(log
(1)
log(
1)lo
g(2)
log(
1)...
log
log(
1)lo
g
log(
1)lo
g(
1).
log
n
kp
pn
kp
p
kp
pn
pp
np
pn
nn
p
11 .
n
kp
kn
kp
FO
RM
UL
A
Pro
of
It is
obv
iou
s th
at,
11
21
1...
.1
n
kp
kn
pn
nk
pp
np
Pro
of
Obv
iou
sly,
1
1 ti
mes
(1)
tim
es
...=
c a
nd
...
=c
.n
nn
nr
kk
rn
nr
cc
cc
cc
cc
cc
Pro
perti
es 7
For
can
d r
¢,
1
1
an
d
.n
nn
nr
kk
r
cc
cc
120
Appl
ied
Mat
hem
atic
s 11
EX
AM
PLE29
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
c.d.
7
5
1 6k
6
3
2k
9
1
(3)
p
4
1
3k
a.c.
Th
e so
luti
ons
of b
. an
d d.
are
left
as
an e
xerc
ise
for
the
stu
den
t.
66
31
4
3
22
216
.k
44
1
33
81.
k
So
luti
on
Pro
of
Obv
iou
sly,
12
31
12
31
tim
es
...
......
.
n
kn
k
nn
nn
kk
n
ca
ca
ca
ca
ca
cc
cc
aa
ac
ca
EX
AM
PLE30
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
c.3 1
1(
)t
t
52
1
2p
p3
1
4k
k
a. b.an
d c.
are
left
as
an e
xerc
ise
for
the
stu
den
t.
33
33
11
44
4(1
23
384.
kk
kk
So
luti
on
Pro
perti
es 8
For
c,
11
.n
nn
kk
kk
ca
ca
Pro
of
1
12
11
1
......
.p
nn
kp
pn
kk
kk
kp
aa
aa
aa
aa
Pro
perti
es 9
For
1 <
p<
n, p
, 1
11
.p
nn
kk
kk
kk
p
aa
a
121
Mat
hem
atic
al In
duct
ion
EX
AM
PLE31
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
17
10
1(1
)p
p
7
4k
k
a.B
y pr
oper
ty, 3
, we
hav
e
Div
idin
g bo
th s
ides
of
this
equ
alit
y by
gi
ves
b.T
he
solu
tion
is le
ft a
s an
exe
rcis
e fo
r th
e st
ude
nt.
7
74
34
1
12
34
56
77!
.1
23
3!k
k
k
kk
k
3
1
,k
k
37
7
11
4
.k
kk
kk
kS
olu
tio
n
EX
AM
PLE32
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
3
1
()
2kk
k3
2
1
2k
k
k
a. b.T
his
par
t is
left
as
an e
xerc
ise
for
the
stu
den
t.
33
32
2
11
1
12
32
22
8
22
(22
2)
(12
3)
29.
kk
kk
k
kk
So
luti
on
Pro
of
1. 2.2
13
11
23
1
23
11
12
3
... ...:
,
0....
nk
n
kk
n
nn
nk
kk
kk
n
aa
aa
ab
bb
bb
aa
aa
ab
bb
bb
b
11
22
33
1
12
31
23
1
()
...
(...
)(
...)
.
n
kk
nn
k
nn
nn
kk
kk
ab
ab
ab
ab
ab
aa
aa
bb
bb
ab
Pro
perti
es 1
0
not
atio
n h
as d
istr
ibu
tive
pro
pert
y ov
er m
ult
iplic
atio
n a
nd
divi
son
.
1. 2.
11
1
(:
):
,
0.n
nn
kk
kk
kk
kk
ab
ab
b
11
()
,n
nn
kk
kk
kk
k
ab
ab
122
Appl
ied
Mat
hem
atic
s 11
Note
not
atio
n d
oes
not
hav
e di
stri
buti
ve p
rope
rty
over
add
itio
n a
nd
subt
ract
ion
i.e.
,
11
1
()
.n
nn
kk
kk
kk
k
ab
ab
Pro
perti
es 1
1R
ule
for
Ch
angi
ng
Bou
nda
ries
For
r, p
,
You
may
wor
k ou
t fo
r th
e pr
oof
by y
ours
elf.
2
.n
pn
pn
kk
pk
pk
kr
pk
rp
aa
a
Pro
perti
es 1
2C
omm
uta
tive
Pro
pert
y of
11
11
()
()
nm
mn
krkr
kr
rk
aa
EX
AM
PLE33
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
c.10
0t
p7
2
1(1
)k
k
10
13
log
.k
k
k
a. b. c.ar
e le
ft a
s an
exe
rcis
e fo
r th
e st
ude
nt.
77
16
22
11
11
(1)
(1) 1
1
1 2
kk
k
kk
kk
2 33 4
4 55 6
61
.7
7
1010
28
12
13
33
21
45
611
log
log
(2)
log
(2)
log
3lo
g4
log
5...
log
10
log
3lo
g4
kk
kk
kk
kk
k
log
4
log
5
log
5
log
6
log1
0...
11lo
g3.
log1
1
So
luti
on
Pro
of
12
31
11 11
121
2122
21
2
12
11
1
12
1 11
()
(...
)
(...
)(
...)
...(
...)
......
(...
)
().
nm
n
krk
kk
kmk
rk
mm
nn
nm
mm
m
rr
nrr
rr
m
rr
nrr m
n
krr
k
aa
aa
a
aa
aa
aa
aa
a
aa
a
aa
a
a
123
Mat
hem
atic
al In
duct
ion
EX
AM
PLE34
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
32
11
()
tp
pt
43
11
2(
)3r k
kr
a. b.T
he
solu
tion
is le
ft a
s an
exe
rcis
e fo
r th
e st
ude
nt.
23
64
34
4
33
11
11
46
24
36
912
30
22
22
2(
)3
33
22
.3
33
33
r kk
kk
rk
k
So
luti
on
EX
AM
PLE35
Eva
luat
e th
e gi
ven
exp
ress
ion
s.
a.b.
110
364
1
log
(2
)k
k
210
(2
)
0
3k
k
a. b.
101
11
1010
1010
33
164
6464
11
31
log
23
1lo
g(
2)
log
2(
3)l
og2
.2
6lo
g2
12
kk
kk
kk
1010
22
20
0
1011
2110
1110
(2
)(
44
)4
411
(2
)38
522
044
209
64
0
33
33
33
.k
k
kk
kk
k
So
luti
on
EX
AM
PLE36
If
then
fin
d th
e va
lue
of n
.5
11
2n k
k
kn
k
Th
e so
luti
on is
left
as
an e
xerc
ise
for
the
stu
den
t.
So
luti
on
Pro
of
Obv
iou
sly,
21
3
12
31
1
...
...
.
kn
n
kn
k
na
aa
aa
k
aa
aa
a
rr
rr
r
rr
Pro
perti
es 1
3
For
r,
1
1
n
kk
k
na
a
k
rr
124
Appl
ied
Mat
hem
atic
s 11
EX
AM
PLE37
Pro
ve t
hat
22
22
(1)
(21)
12
3...
.6
nn
nn
EX
AM
PLE38
Pro
ve t
hat
22
23
2(
1)1
23
...(
).
2n
nn
Rem
embe
r th
at w
e h
ave
prov
en
this
for
mu
la b
efor
e. B
ut
now
,
we
will
pro
ve i
t by
cal
cula
tin
g
the
area
of
the
rect
angl
e h
avin
g
the
side
s 1
+ 2
+ 3
+ .
.. +
n
and
2n+
1 u
nit
s. I
n F
igu
re 1
.2,
the
lett
ers
A1,
A2,
A3,
...,
An
den
ote
the
area
s pa
inte
d by
th
e
sam
e co
lor.
A(A
BC
D)
= A
1+
A2
+ A
3+
. . .
+ A
n
(1 +
2 +
3 +
. . .
+ n
) (2
n+
1)
= 3
+ 1
2 +
27
+ .
. . +
3n
2
Hen
ce
2
1
(1)
(21)
.6
n k
nn
nk
22
1
(1)
(21)
3(1
49
...3
2
n k
nn
nn
k
So
luti
on
nBC
AD
32
1
12222
A3
A2
A1
n � times
An
By
the
Bin
omia
l th
eore
m t
hat
(x+
)4
= x
4+
4x3 y
+ 6
x2 y2+
4xy
3+
y4 .
Usi
ng
this
exp
ress
ion
yie
lds
(1 +
1)4
= 1
4+
4 ·
13· 1
+ 6
· 12
· 12
+ 4
· 1
· 13
+ 1
4
(2 +
1)4
= 2
4+
4 ·
23· 1
+ 6
· 22
· 12
+ 4
· 2
· 13
+ 1
4
(3 +
1)4
= 3
4+
4 ·
33· 1
+
6 · 3
2· 1
2+
4 ·
3 · 1
3+
14
......
......
......
......
......
......
......
......
......
......
......
......
......
.....
......
......
......
......
......
......
......
......
......
......
......
......
......
.....
(n+
1)4
= n
4+
4 ·
n3
· 1 +
6 ·
n2
· 12
+ 4
· n
· 13
+ 1
4
So
luti
on
125
Mat
hem
atic
al In
duct
ion
EX
AM
PLE40
Pro
ve t
hat
by u
sin
g su
mm
atio
n n
otat
ion
.2
31
47
...3
22
nn
n
1 11
2 2
14
7...
32
(32)
32
(1)
32
2
33
42
3.
2
n k nn
kk
nk k nn
n
nn
n
nn
So
luti
on
Add
ing
thes
e eq
ual
itie
s si
de b
y si
de g
ives
(n+
1)4
= 1
4+
4(1
3+
23
+ ..
. + n
3 ) +
6(1
2+
22
+ ..
. + n
2 ) +
4(1
+ 2
+ ..
. +n
) +
n.
Sin
ce
we
get
By
rear
ran
gin
g th
is e
qual
ity,
we
will
hav
e
Hen
ce,
22
3
1
2
(1)
4
(1)
()
.2
n k
nn
k
nn
34
32
22
1
22
42
(2
1)
(1)
.
n k
kn
nn
nn
n
nn
43
23
1
(1)
(21)
(1)
46
44
64
62
n k
nn
nn
nn
nn
nk
n
2
11
(1)
(21)
(1)
an
d
62
nn
kk
nn
nn
nk
k
43
23
2
11
1
46
41
14
64
nn
n
kk
k
nn
nn
kk
kn
EX
AM
PLE39
Fin
d th
e su
m –
11 –
7 –
3 +
1 +
5 +
... +
63.
Th
e so
luti
on is
left
as
an e
xerc
ise
for
the
stu
den
t.
So
luti
on
126
Appl
ied
Mat
hem
atic
s 11
EX
AM
PLE41
Fin
d th
e su
m3
2
1
(46
2).
n k
kk
k
= n
2 (n+
1)2
– n
(n+
1)(
2n+
1)
+ n
(n+
1)
= n
(n+
1)[
n(n
+ 1
) –
(2n
+ 1
) +
1]
= n
(n+
1)(
n2
+ n
– 2n
– 1
+ 1
)
= n
(n+
1)(
n2
– n
)
= n
4–
n2 .
32
2
1
(1)
(1)
(21)
(1)
(46
2)
4(
)6
22
62
n k
nn
nn
nn
nk
kk
So
luti
on
EX
AM
PLE42
Pro
ve t
hat
(1)
(2)
12
23
34
...(
1).
3n
nn
nn
2
11
12
23
34
...(
1)(
1)(
)
(1)
(21)
(1)
62
(1)
21
()
(1)
23
(1)
24
23
(1)
(2)
.3
nn
kk
nn
kk
kk
nn
nn
n
nn
n
nn
n
nn
n
So
luti
on
EX
AM
PLE43
Eva
luat
e10
119
13
2
(2
(3)
):6.
kk
k
Sin
ce
wh
ere
n–
p+
1 is
th
e n
um
ber
of t
erm
s, k
[p, n
], w
e ge
t
1011
9
13
2
(2
(3)
):6
((2
10)
(3
9)):
(612
)
2027
15.
722
kk
k
()
(1)
n
kp
cc
np
So
luti
on
127
Mat
hem
atic
al In
duct
ion
EX
AM
PLE44
Pro
ve th
at
by u
sin
g su
mm
atio
n n
otat
ion
.2
44
44
(1)
(21)
(33
1)1
23
...30
nn
nn
nn
Th
e so
luti
on is
left
as
an e
xerc
ise
for
the
stu
den
t.S
olu
tio
n
EX
AM
PLE45
Fin
d th
e su
m
102
3
(4
).k
kk
1010
22
2
33
2
2
(4
)((
2)4(
2))
(4
kk
kk
kk
kk
44k
88
2
11
8)(
4)
89
174
86
2032
172.
kk
k
So
luti
on
EX
AM
PLE46
Eva
luat
e 4
2
0
(23
7).
k
kk
44
12
2
00
1
52
1
(23
7)[2
(1)
3(1)
7]
(28)
56
115
62
58
62
110
1540
55.
kK K
kk
KK
kk
So
luti
on
EX
AM
PLE47
Eva
luat
e 15
2
4
2.k
k
1515
153
122
31
4
44
43
1
1216
432
84
24
44
4
14
44
22
4.
14
33
kk
kk
kk
kk
So
luti
on
128
Appl
ied
Mat
hem
atic
s 11
EX
AM
PLE48
Eva
luat
e 20
1
1
1(
1)(
).
2n
n
k
Th
e so
luti
on is
left
as
an e
xerc
ise
for
the
stu
den
t.
So
luti
on
EX
AM
PLE50
Eva
luat
e 40
1
(2
12
1).
k
kk
40
1
(2
12
1)3
15
37
5...
8179
811
91
8.
k
kk
So
luti
on
EX
AM
PLE51
Eva
luat
e k
p
kp
44
11
((2
35)
).
kp
kk
kp
kk
44
44
11
11
45
((2
35)
)(2
43
54)
(810
)2
45
810
480
4012
0.2
So
luti
on
EX
AM
PLE49
Fin
d th
e su
m
71
6
1.
(1)
kk
k
Fir
st w
ay:
Usi
ng
the
“For
mu
la 1
.5”,
we
get
Sec
ond
way
:
Sin
ce
we
obta
in
7171
66
11
1(
)(
1)1
11
11
11
...6
77
871
72
11
11.
672
72
kk
kk
kk
11
1,
(1)
1k
kk
k
7171
5
61
1
11
1(
1)(
1)(
1)
715
11.
726
72
kk
kk
kk
kk
k
So
luti
on
129
Mat
hem
atic
al In
duct
ion
EX
AM
PLE52
Eva
luat
e k
k
30
21
2.
41
Sin
ce
we
get
kk
kk
k
3030
21
1
21
1(
)2
122
14
1
11
11
1(1
...)
33
559
61
160
(1)
.61
61
kk
k221
1(
),2
12
14
1S
olu
tio
n
EX
AM
PLE53
Eva
luat
e k
kk
20
20
2.
32
kk
kk
kk
kk
k
2020
20
20
00
21
11
22
()
(1)
(2)
12
32
11
11
11
12
(1...
)2
23
34
2122
121
2(1
).
2211
So
luti
on
EX
AM
PLE54
Eva
luat
e a
b
ba
34
00
(3)
.
ab
a a
ba
a a
34
3
00
0 3 0
45
(3)
((4
1)3
(41)
)2
(25
5)
34
25(3
1)5
2
100
3070
.
So
luti
on
EX
AM
PLE55
Let
x1
and
x 2be
th
e ro
ots
of t
he
quad
rati
c eq
uat
ion
x2
– 6x
+ 8
= 0
pro
vide
d th
at x
1<
x2.
If f
(x)
= 4
x–
1 fi
nd
the
sum
i
ii
xf
x2 1
().
Sin
ce t
he
root
s of
th
e eq
uat
ion
x2
– 6x
+ 8
are
x1
= 2
an
d x 2
= 4
, we
hav
e
= x
1f(
x 1)
+ x
2f(
x 2)
= f
(2)
+ 4
· f(
4)
=
2 · (
4 · 2
– 1
) +
4 ·
(4 ·
4 –
1) =
14
+ 6
0 =
74.
ii
i
xf
x2 1
()
So
luti
on
130
Appl
ied
Mat
hem
atic
s 11
EX
AM
PLE56
If
fin
t th
e su
m
a t
t3
1
.k
kk
a5
2
0
(23
5)5
,
Th
at is
, 5a
= 3
5
a=
7. H
ence
,
t
t7
32
2
1
78
()
2878
4.2
k
kk
52
0
56
115
6(2
35)
23
56
62
110
4530
35.
So
luti
on
EX
AM
PLE57
If
and
then
fin
d f
o g(
5)x p
gx
p2
1
()
,k
kk
52
0
(23
Sin
ce f
o g(
5) =
f(g
(5))
, we
obat
ain
pk
fg
fp
ff
k5
552
11
56
1155
56(
(5))
()
()
(55)
1540
.6
2
So
luti
on
EX
AM
PLE58
Pro
ve t
hat
n p
pp
n1
!(
1)!
1.
= 2
! –
1!+
3!
– 2!
+ 4
! –
3! +
... +
(n
+ 1
)! –
n!
= (
n+
1)!
– 1
.
nn
pp n
n
pp
pp
pp
pp
pp
p
11 1
1
!(
11)
!
((1)
!!)
((1)
!!)
So
luti
on
EX
AM
PLE59
Pro
ve t
hat
n k
kk
n1
11
.(
1)!
(1)
!
nn
kk n k n k
kk
kk k k
k
kk
nn
n
11 1 1
11
(1)
!(
1)! 1
1(
)(
1)!
(1)
!
11
()
!(
1)!
11
11
11
1...
1!2!
2!3!
4!!
(1)
!
11
(1)
!
So
luti
on
131
Mat
hem
atic
al In
duct
ion
EX
AM
PLE60
Eva
luat
e k
k
99
1
1(1
). 1
kk
kk
k
9999
11
1(1
) 11
99!
99!
1.
100!
99!
100
100
So
luti
on
EX
AM
PLE61
Eva
luat
e k
k
k62
4 5
log
(1)
.
= lo
g 56
log 6
7 lo
g 78
... lo
g 624
625
= lo
g 562
5 =
log 5
54=
4.
kk
k62
4 5
log
(1)
So
luti
on
EX
AM
PLE62
Pro
ve t
hat
n
k
nn
k22
11
(1)
.2
nn
kk
nn
n
kk
k
kk
k kk
kk
kk
kk
nn
nn
nn
nn
2
22
22 2
22
11
(1)
(1)
(1)
11
()
()
12
31
34
51
(...
)(
...)
23
42
34
11
1.
22
So
luti
on
EX
AM
PLE63
Fin
d th
e va
lue
of x
, if
tx
t92
25
1
316
()
()
.4
9
By
usi
ng
“Pro
pert
y 7”
, we
get
Th
is g
ives
x=
–25
.
t
tt
xx
x
t
xx
9 1
92
22
52
59
104
10
1
904
10
316
316
34
()
()
()
()
()
()
49
49
43
33
()
()
904
10.
44
So
luti
on
132
Appl
ied
Mat
hem
atic
s 11
EX
AM
PLE66
Eva
luat
e k
p
kp
43
11
().
kp
k kk
kp
k
kk
43
43
11
1
44
34
34
313
7
11
()
(3!
)
66
6(4
!)2
3.
So
luti
on
EX
AM
PLE65
Fin
d th
e va
lue
of x
, if
xt
x
t
13
3
2
48
.
x2+
3x
= 9
x–
9
x2+
3x
– 9x
+ 9
= 0
x2–
6x+
9 =
0
(x–
3)2
= 0
Hen
ce, w
e co
ncl
ude
th
at x
= 3
.
x t
xx
tt
xt
xx
tt
xx
xx
xx
xx
1
1(
1)3
31
33
33
21 (
1)3
3(
1)2
99
2
48
48
48
48
22
So
luti
on
EX
AM
PLE67
Eva
luat
e i
j
i j
34
11
().
ij
ii
ii
ij
44
34
33
11
11
4 3
()
()
4!24
(3!)
3.
3224
So
luti
on
EX
AM
PLE68
Eva
luat
e a
b
ab
102
11
().
aa
ba
ab
aa
aa
102
1010
2
11
11
((
))(
1)(
2)(
32)
1011
2110
113
210
62
385
165
2057
0.
So
luti
on
EX
AM
PLE64
Eva
luat
e k
kk
kk
25
21
32
.4
kk
kk
kk
kk
kk
kk
kk
kk
25
55
5
21
11
1
32
(1)
(2)
12
(4)
44
23
45
63
46
71
()
()
6(
)1.
12
34
55
68
96
So
luti
on
133
Mat
hem
atic
al In
duct
ion
EX
AM
PLE69
Eva
luat
e n
nk
n2
100 1
1
((
1)).
n
nk
k kk
n2
210
0 11
11
11
1
22
2
11
50 ti
mes
((
1))
(1
23
45
6...
9910
0)
(11
1...
1)50
5025
00.
So
luti
on
EX
AM
PLE70
Eva
luat
e n
nk
n
kk
15
12
1(l
og).
n
nn
nn
nk
nn
nn
nn
kn
kn
nn
n
1515
12
2
15
2
15
2
23
415
2
12
34
1(
(log
)(l
oglo
glo
g...
log
)1
23
23
41
(log
...)
12
3
log
(1)
log
3lo
g4
log
5...
log
16
log
16
4.
So
luti
on
EX
AM
PLE71
If
fin
d f
o g(
4)x
xx
x
mt
kn
mf
xg
xn
21
11
11
1(
) a
nd
(
),
612
Sin
ce f
og (
4) =
f(g
(4)
), w
e ob
tain
mk k
kmf
42
11
44
11
4
(2)
6
31
62
11
()
.2
16tn
t
fog
fn
f ff3
4 11
4 1
1(4
)(
)12 1
(6)
12 24 ()
(2)
12
So
luti
on
134
Appl
ied
Mat
hem
atic
s 11
EX
ER
CIS
ES
3.2
3.E
valu
ate
the
indi
cate
d su
ms
belo
w.
a.b.
c.d.
e.f.
k
k26
7 87
sin
(3)
k
k35
9 180
cos
n
k
n30
1
(1)
sin
2p
p
p7
0
(1)
(21)
k
k
k3
12
2
(1)
k
k
a21
2
1
(1)
2.E
valu
ate
the
indi
cate
d su
ms
belo
w.
a.b.
c.d.
e.f.
g.h.
i.j.
k.l.
k
k
a11
3
log
k
kk
3
0
(1)
!
mk
kk
2010
10
(3)
(21)
n
kn
k2
(21)
p
p7
3
(31)
p
p1
2
7
(216
)
t
t9
2
1
(32)
k
kk
122
0
(1)
k
k7
2
0a
a17
0
(21)
k
k40
3
(2)
k
k50
1
1.E
valu
ate
the
indi
cate
d su
ms
belo
w.
a.b.
c.
d.e.
f.
g.h.
i.n
k2
2
(1)
k
n4
1
(1)
k
2
0
r
e5
1t
3
2
10m
8
2
(6)
p22
3
7k10
0
(2)
k12
1
3
gh.
i.j.
k.l.
m.
n.
o.p.
q.r.
s.t.
u.v.
w.
x.8
82
2
22
(3
)(3
1)k
k
kk
k
77
2
11
(2
)(2
4)k
k
kk
k
1313
1
11
22
xx
xx
133
2
13
(23
)k
kk
k
52
5
()
r
kk
43
4k
k
50
1(
1)!
k
kk
100 1
!k
kk
79
32
1lo
g(1
) 1p
p
100 1
1lo
g1
kk
89
1
(si
nco
sk
kk
2
(1
1n
k
kk
22
42
32
23
nn
ki
ki
ki
50
32
2k
kk
k
50
23
1 32
pp
p
9
21
23
3k
kk
k
k
ii
1000
2
1
()
(1)
k
k90
2
1
sin
4.E
valu
ate
the
indi
cate
d su
ms
belo
w.
a.b.
c.d.
e.f.
2
11
1
kr
pr
t
p3
11
12
kn
kn
m
t
15
12
(3)
ab
aab
4
01
(22)
km
km
54
2
10
(3)
kn
nkn
75
10
(2)
3m
n
nm
A.
Su
mm
ati
on
No
tati
on
135
Mat
hem
atic
al In
duct
ion
6.P
rove
th
e fo
llow
ing
expr
essi
ons.
a.b.
c.d.
0
1n
k
rk
rn
kn
2
0
2n
k
nn
kn
1
0
(1)
(2)
2n
n
k
nk
nk
1
1
2n
n
k
nk
nk
5.P
rove
th
at
0
1co
tco
t2.
2si
n2
nn
nk
aa
a
8.If
fi
nd
the
sum
6
1
3.
kk
a6
1
(23)
58,
kk
a
9.If
fi
nd
n.
0
(23)
63,
n
k
k
10.
If
fin
d n
.(
1)
1
log
39
log
27,
nk
k
19.
If
fin
d a 6
.6
7
21
3
(23)
2,
kk
kk
aa
20.
If
fin
d th
e va
lue
of
a+
b+
c.
2
1
(62)
2
n k
anbn
ck
12.
If
fin
d n
.2
92
2,1
nn
kk
17.
If
fin
d a 4
+ a
5.2
1
1
22
,n
nk
k
a
18.
If a
1=
1 a
nd
a n=
an
–1+
2, f
ind
the
sum
30
1
.k
k
a
13.
If
fin
d f(
i) w
her
e i2
= –
1.
4
0
()
,n
k
k
fx
x
15.
If f(
x) =
x+
1, x
1=
1, x
2=
2 an
d
fin
d f(
i) w
her
e i2
= –
1.
32
1
()
50,
ii
i
xf
x
16.
Let
x1an
d x 2
are
the
root
s of
the
quad
ratic
equ
atio
n
x2+
2x
– 3
= 0
. If f
: +
+, f
(x)
=2x
+ 3
then
eval
uate
2
11
().
ii
xf
x
11.
If
fin
d a.
204
1
1
12
2,
ka
k
14.
Let
f :
+
+,
Fin
d th
e va
lue
of f
–1(8
)1
()
21.
x k
fx
k
7.E
valu
ate
the
indi
cate
d su
ms
belo
w.
a.b.
c.d.
e.f.
64
21
64lo
g(
)k
kk
210
0
10
kk
10
5
102k
kk
0
nk
k
nx
k
60
0
5
k
k
k
6
0
7
kk
136
Appl
ied
Mat
hem
atic
s 11
26.
If f
(x)
= 2
x+
3 a
nd
g(x)
= 3
x–
2 fi
nd
the
sum
51
1
()
().
x
fog
x
29.
If
fin
d x.
33
1
()
30,
x
km
x
mk
30.
If
fin
d a.
33 3
(24)
168,
aa
ka
k
k
32.
If
and
fin
d f(
2).
1 11
()
2,p
x pt
gx
2
11
()
((
))
xk
km
fx
gm
m
31.
If x
1=
2, x
n=
x2 n
– 1
–1 a
nd
f(x)
= x
+ 1
th
en fi
nd
the
sum
3
3
11
()
().
ji
ij
i
xx
fx
27.
If x
1=
2, x
2=
4 an
d f(
x) =
2x , g(
y) =
2lo
g 4y
then
fin
d th
e su
m
2 1
()(
). ii
gof
x
28.
Let
f:
+, f
(x)
= lo
gan
d
g:
N+
, g(
x) =
log
then
fin
d th
e su
m
50
3
[(
)(
1)].
k
fk
gk
1(1
)x
1(1
)x
21.
If
fin
d th
e
valu
e of
x–
2y.
23
11
(3
)12
an
d
()
3,k
p
kxy
pxy
23.
If
fin
d th
e su
m 2
5
+ 4
7
+ .
..
+ 3
0 33
. in
ter
ms
of A
.
152
1
(42
)k
Ak
k
25.
If f
(x)
= x
– 3
and
g(x)
= x
2 , th
en f
ind
the
sum
5
1
()(
).x
gof
x
24.
If
and
then
fin
d
in t
erm
s of
Aan
d B
.
1
12
1
n kk
2
1
1,
n
kn
Bk
1
124
n kk
22.
Let
x1an
d x 2
be th
e ro
ots
of th
e qu
adra
tic e
quat
ion
x2+
(2a
+ b
)x+
a–
b=
0.
Fin
d th
e va
lue
of a
(ab)
, if
22
2
11
22
an
d
5k
kk
k
xx
137
Mat
hem
atic
al In
duct
ion
36.
If
then
fin
d th
e va
lue
of x
.1
1
381
,x
kx
k
37.
If
then
fin
d th
e va
lue
of x
.4
12
1lo
g(1
) 1
1
32
,6
xp
p
38.
If
fin
d x.
22
log
(1)
log
(25)
,x
kk
kx
39.
If
fin
d n
.3
(1)
1
212
8,n
k
k
41.
Eva
luat
e if
f(x
) =
3x
+ 2
.6
1
1
((
)1)
,k
fk
42.
If f
(x)
= x
, th
en f
ind
wh
ere
i2=
–1.
100 1
[(
)]k
k
fi
40.
Let
F
ind
the
smal
lest
val
ue
of n
wh
ich
mak
es A
zero
.
32
1
(7
12).
n k
kk
k
B.
Mu
ltip
licati
on
No
tati
on
33.
Eva
luat
e th
e gi
ven
exp
ress
ion
s be
low
.
a.b.
c.
d.e.
f.
g.h.
i.7
3
(21)
n
k5
1
(2)
k
n7
2p
e
4
0
3(
)2
k
3
1
2r
4
2
3 ()
4k
7 3
1 ()
3t
5
0
(1)
k
4
1
5k
34.
Eva
luat
e th
e gi
ven
exp
ress
ion
s be
low
.
a.b.
c.d.
e.f.
g.h.
i.j.
k.l.
m.
n.
o.p.
q.r.
s.t.
u. x. w.
5050
22
01
(3
2):
(2
)k
k
kk
kk
2020
11
11
log
(1)
log
(1)
aa
aa
33
2
10
(1)
(1)
xx
xx
x
89
1
tan
k
k3
72
8
3kk
k
10
22
44
(1)
k
k k
275
11
k
k k
250
2
1
(4
12)
k
kk
26
()
2
2k
k
k
k
25
21
log
4k
k
26
()
2
2k
k
k
k
50!
1
log
3kk
k
21
3
1(1
)2
rr
154
log(
1) 1
3
2p
p
503
1
9k
kk
28
2
log
(1)
kk
k3
2
1
log
2m
am
14
13k
k
k5
1
2k
k
k
62
1
2
3t
k
7 0
2t
t
7
3
(36)
r
r41
1k
k
35.
Eva
luat
e th
e gi
ven
exp
ress
ion
s be
low
.
a.b.
c.d.
24
3
11
ji
i j
180
180
sin
cos
10
(2)
kk
kk
45
21
()
rn
nr3
31
(1)
k
kn
138
Appl
ied
Mat
hem
atic
s 11
46.
If
then
fin
d n
.3
8
11
39
2,
k
kk
kn
48.
If
and
fin
d g(
63).
() 1
1(
)(1
)f
x
p
gx
p1
1(
)1
x k
fx
kk
50.
If
and
fin
d
go
f(3
).
2
11
()
(1)
,x
xk
kk
gx
k2
1
11
()
2x
xk
kk
fx
49.
If
then
fin
d 2
log
(1)
.a
kk
a
k
7
2
(2)
(3)
(4)
k
ak
kk
51.
If
and
x 1=
2, x
2=
4, f
ind
the
sum
2 1
().
ii
i
xf
x
1
1
()
2x
k
k
fx
52.
Let
x1an
d x 2
be th
e ro
ots
of th
e qu
adra
tic e
quat
ion
x2+
(a
+ b
)x+
b–
2a=
0.
If
and
then
fin
d th
e va
lue
of t
he
pair
(a,
b).
2 1
6i
i
x
2 1
5i
i
x
53.
Eva
luat
e if
f
: +
,
and
g :
++, g
(x)
= x
.1
()
fx
x
11
11
()
(),
kn
k
kr
gk
fr
54.
Let
x 1
and
x 2be
th
e ro
ots
of
the
equ
atio
n
x2–
5x–
n=
0. I
f fi
nd
the
sum
of
the
valu
es w
hic
h n
can
tak
e.
22
1
11
(())
0k
ki
ik
i
xx
47.
If
then
fin
d 4
2
1k
k
x1
21,
n
kk
xn
45.
Eva
luat
e th
e gi
ven
exp
ress
ion
s be
low
.
a.b.
c.d.
e.f.
5
3
2(2
4)
1
2k
xk
k
710
11
(1)
(2)
kk
kp
57
11
1(1
)k
kk
510
11
rk
r k
107
11
2k
p
k3
5
11
2r
k
44.
If
and
fin
d (g
o f)
(3).
1 1
()
,x k
gx
k1
()
2x
k
k
fx
43.
If x
1=
1,
x 2=
2,
x 3=
3 an
d f(
x) =
2x
– 1
then
eval
uat
e 3
21
.[
()]
i
iix
fx
139
Cha
pter
Rev
iew
Tes
t 3A
CH
AP
TE
R R
EV
IE
W T
ES
T 3
A
5.If
th
en w
hat
is t
he
valu
e of
th
e su
m
2n+
(2n
+ 1
) +
(2n
+ 2
) +
... +
3n
in te
rms
of M
?
A)
M2
+M
B)
M+
1 C
)3M
D)
4ME
)5M
1
n k
kM
2.T
he
sum
of
the
cube
s of
th
e n
um
bers
fro
m 1
to
n
is K
= 1
3 +
23
+ 3
3 +
... +
n3 . H
ow m
uch
will
K
incr
ease
if
each
of
the
nu
mbe
rs 1
, 2,
3,
...,
nis
incr
ease
d by
1?
A)
nB
)n
3C
)n
(n2
+3n
)
D)
n(n
2+
3n
+ 3
)
E
)2
(1)
()
2n
n
3.W
hat
is t
he
valu
e of
th
e su
m
m2
+ (
m+
1)2
+ (
m+
2)2
+ ..
. + (
2m)2
wh
ere
m
is a
nat
ura
l nu
mbe
r?
A)
B)
C)
D)
E)
(1)
(14
3)3
mm
m
2(
1)(1
21)
3m
mm
(1)
(21)
12m
mm
(1)
(18
1)6
mm
m(
1)(1
41)
6m
mm
6.If
th
en w
hat
is
the
valu
e of
n?
A)
9 B
)10
C
)11
D)
12
E)
13
33
33
33
33
13
5...
(21)
199
242
24
6...
(2)
nn
7.If
th
en t
he
valu
e of
xis
A)
28
B)
25
C)
23D
)20
E
)19
14
1515x
kk
kk
8.W
hat
is t
he
valu
e of
th
e su
m
A)
233B
)234
C)
235D
)236
E)
237
32
1
32!
wh
ere
?(
)!!
k
nn
kk
kn
kk
10.
Wh
at is
th
e va
lue
of t
he
sum
A)
B)
C)
D)
E)
11 99 22
11 139 11
7 9
120 4
1?
1(
1)k
kk
kk
9.T
he
con
secu
tive
od
d in
tege
rs
are
grou
ped
as
follo
ws
1; (
3, 5
); (
7, 9
, 11)
; (13
, 15,
17,
19)
; ...
Wha
t
is t
he
sum
of
the
nu
mbe
rs in
th
e 10
’th g
rou
p?
A)
760
B)
897
C)
981
D)
1000
E
)10
21
4.Tr
ian
gula
r n
um
bers
can
be
repr
esen
ted
by d
ots
that
are
arr
ange
d in
th
e sh
ape
of a
n e
quila
tera
l
tria
ngl
e, a
s sh
own
bel
ow. T
he
firs
t fo
ur
tria
ngu
lar
nu
mbe
rs a
re g
iven
.
Wh
at is
th
e 15
’the
tria
ngu
lar
nu
mbe
r?
A)
3B
)5
C)
–5
D)
10
E)
2
t 1 =
1t 4
= 1
0t 2
= 3
t 3 =
6
1.If
th
e su
m o
f th
e fi
rst
3npo
siti
ve i
nte
gers
is
150
mor
e th
an t
he
sum
of t
he
firs
t n
pos
itiv
e in
tege
rs,
then
wha
t is
the
sum
of t
he fi
rst 4
n po
sitiv
e in
tege
rs.
A)
270
B)
300
C)
330
D)
360
E)
390
140
Appl
ied
Mat
hem
atic
s 11
13.
If
then
the
valu
e of
A–
Bis
A)
0 B
)C
)D
)E
)14
313
212
610
711
110
126 41
120
20
111
(1)
1 a
nd
n
nn
AB
nn
17.
If
then
th
e va
lue
of x
is
A)
3 B
)4
C)
5D
)6
E)
7
1 1
3x n
xn
x
18.
Wh
at is
th
e va
lue
of t
he
sum
A)
328B
)329
C)
330D
)331
E)
332
27
1
(21)
3?
n
n
n
20.
Def
ine
(na)
! fo
r n
and
a po
siti
ve t
o be
wh
ere
mis
th
e gr
eate
st i
nte
-
ger
for
wh
ich
n>
ma.
Th
en t
he
quot
ien
t
is e
qual
to
A)
210B
)212
C)
216D
)218
E)
220
8 2
(72
)!(1
8)!
0
()!
()
m
ak
nn
ka
19.
Wh
at is
th
e va
lue
of t
he
sum
26 +
37
+ 5
0 +
65
+ ..
. + 4
01?
A)
2796
B
)28
18
C)
2848
D)
2856
E
)28
72
15.
Wh
at is
th
e va
lue
of t
he
prod
uct
A)
B)
C)
D)
E)
37 2531 29
27 2523 25
17 1544
44
(1)(
1)(
1).
..(1
)?4
925
625
14.
Wh
at is
th
e va
lue
of t
he
prod
uct
A)
B)
C)
D)
E)
2047
2047
a10
2410
23a
2047
2048
a10
2310
24a
511
512
a
1024
48
...?
aa
aa
12.
Con
side
r th
e tr
ian
gula
r ar
ray
form
ed b
y ro
tati
ng
the
prod
uct
in
a s
tan
dard
mu
ltip
licat
ion
tab
le,
that
is,
Wh
at i
s th
e su
m o
f th
e el
emen
ts i
n t
he
18› th
row
?
A)
1020
B)
1060
C
)11
00
D)
1140
E
)11
80
1
22
34
3
46
64
58
98
5
......
......
......
......
......
......
......
......
11.
Wh
at is
th
e va
lue
of t
he
sum
A)
B)
C)
D)
E)
18 9
18 8
18 7
17 8
17 6
10
0
71
wh
ere
?
11
n
nn
nn
nr
rr
16.
Wh
at is
th
e va
lue
of t
he
sum
A)
B)
C)
D)
E)
1516
15
(41)
(41)
303
4
1516
15
(21)
(21)
303
2
1516
15
(41)
(41)
303
4
1516
15
(41)
(41)
303
4
1516
15
(21)
(21)
303
2
15
21
1(2
?2
)n
nn
141
Cha
pter
Rev
iew
Tes
t 3A
21.
If i
= ò
–1,
then
th
e su
m
equ
als
A)
B)
–10iñ
2 C
)
D)
E)
2(2
120
)2
i2
(21
20)
2i
212
22 2
40
0
cos(
4590
)n
n
in
23.
If
for
k=
1, 2
, ...,
n–
1 an
d
x 1=
1 t
he
valu
e of
th
e su
m
equ
als
A)
B)
C)
D)
E)
23
4n
n2
4n
n
21
2n
32
n1
2n
1
n
kk
x
1
1 2k
kx
x
28.
Wh
at is
th
e va
lue
of t
he
sum
A)
1075
log
2B
)13
75 lo
g 2
C)
1825
log
2D
)20
25 lo
g 2
E)
2175
log
2
452
1
1
log(
2ta
n(2
1))?
n
n
n
30.
If
then
wh
at is
th
e va
lue
of a
+ b
+ c
?
A)
189
B)
200
C)
216
D)
225
E)
256
53
43
2
3n
kn
kan
bncn
26.
Th
e gr
aph
of
the
fun
ctio
n f
(x)
is g
iven
bel
ow.
Wh
at is
th
e va
lue
of t
he
sum
A)
–11
B)
–10
C)
–8
D)
–6
E)
–4
3 2 11
2
�2
�1
0
�2y
x
f(x
)
2
2
()?
x
xf
x
22.
Wh
at is
th
e va
lue
of t
he
sum
1 +
(1
+ 2
) +
(1
+ 2
+ 3
) +
... +
(1
+ 2
+ 3
+
... +
20)
?
A)
1750
B)
1540
C
)11
90
D)
970
E)
875
24.
Th
e sm
alle
st v
alu
e of
n s
uch
th
at t
he
proc
ut
exce
eds
100,
000
is
A)
7 B
)8
C)
9D
)10
E
)11
11
1
10k
n k29
.If
x1
= –
3, x
2=
1 a
nd
f(x)
= x
2–
4 th
en w
hat
is
the
valu
e of
th
e su
m
A)
–15
B)
–10
C)
–5D
)0
E)
5
2
1
(4)
()?
nn
n
xf
x
25.
Th
e ro
ots
of th
e eq
uat
ion
x3 +
4ax2 +
bx+
2 –
3a=
0
are
x 1,
x 2an
d x 3
. If
th
en t
he
valu
e of
a is
A)
B)
C)
D)
E)
5 44 5
4 72 5
4 5
33
11
(2)
nn
nn
xx
27.
Wh
at is
th
e va
lue
of t
he
sum
A)
2B
)3
C)
4 D
)5
E)
6
24
1
1?
21
21
nn
n
142
Appl
ied
Mat
hem
atic
s 11
CH
AP
TE
R R
EV
IE
W T
ES
T 3
B
7.W
hic
h o
ne
of t
he
follo
win
g is
equ
al t
o
A)
4n·
(n3 )!
B)
((4n
)!)3
C)
64(n
!)3
D)
22n(n
!)3
E)
4 · (
n!)
3
3
1
(4)?
n k
k
4.W
hic
h o
ne
of t
he
follo
win
g ex
pres
s th
e su
m
2 +
7 +
12
+ ..
. + 1
02?
A)
B)
C)
D)
E)
21
1
(31)
n
n10
2 1
(53)
n
n
102 1
(2)
n
n21
1
(53)
n
n21
1
(52)
n
n
8.W
hic
h o
ne
of t
he
follo
win
g is
equ
al
A)
42n
2+
8n
– 1
B)
12n
2–
11n
– 1
C)
42n
2–
3nD
)42
n2
– 3n
– 1
E)
42n
2+
11n
– 1
5
2
(41)
?n
kn
k
5.W
hic
h o
ne
of t
he
follo
win
g ex
pres
s th
e su
m
1 · 3
+ 3
· 7
+ 5
· 11
+ ..
. + 2
1 · 4
3?
A)
B)
C)
D)
E)
112
1
(2)
n
nn
112
1
(87
2)n
nn
432
1
(32
2)n
nn
212
1
(46
5)n
nn
112
1
(86
1)n
nn
2.W
hat
is th
e re
sult
of t
he
prod
uct
A)
0 B
)1
C)
log
(tan
87°)
D)
E)
1lo
g(ta
n3
)27
29!
log(
tan
)2
3
29
2
log(
tan
3)?
k
k
1.a
= 1
+ 2
+ 3
+ ..
. + 2
3
b=
12
+ 2
2 +
32
+ ..
. + 2
32
c=
13
+ 2
3 +
33
+ ..
. + 2
33
is g
iven
.
wh
at is
th
e va
lue
of t
he
sum
1 · 2
· 3
+ 2
· 3
· 4 +
3 ·
4 · 5
+ ..
. + 2
3 · 2
4 · 2
5
in t
erm
s of
a, b
, an
d c?
A)
a+
3b
+ 2
cB
)a
+ 2
b+
3c
C)
3a+
2b
+ c
D)
2a+
3b
+ c
E)
3a+
b+
2c
9.f
: +
+, g
: +
+
are
give
n.
Wh
at is
th
e va
lue
of f
og (
4)
A)
3501
B)
3511
C)
3526
D)
3541
E)
3561
11
()
3 a
nd
(
)!
xx
k
kk
fx
gx
k
6.
is g
iven
.
Th
e va
lue
of B
– A
is
A)
390
B)
410
C)
420
D)
430
E)
450
1111
1111
11
11
23
4...
11k
kk
k
B
23
411
11
11
...k
kk
k
Ak
kk
k
3.
(n
N+)
is g
iven
.
Wh
ich
on
e of
th
e fo
llow
ing
is e
qual
to
the
sum
(m+
1)
(m+
2)
+ (
m+
3)
+ ..
. (2m
– 1)
(2m
)?
A)
B)
C)
D)
E)
(1)
(2)
3m
mm
(1)
(1)
3m
mm
(1)
(1)
(23)
3m
mm
(1)
(74)
3m
mm
(1)
(74)
3m
mm
(1)
(2)
12
23
34
...(
1)3
nn
nn
n
143
Cha
pter
Rev
iew
Tes
t 3B
11.
If
then
th
e va
lue
of n
is
A)
5B
)7
C)
8 D
)9
E)
10
22
46
...2
1024
13
5...
(21)
nn
nn
12.
If
then
wh
ich
on
e of
the
follo
win
g is
equ
al t
o 4a
– 2b
+ c
?
A)
476
B)
493
C)
507
D)
521
E)
546
132
2
1
()
,k
xk
axbx
c
15.
Wh
at is
th
e re
sult
of
sum
A)
7B
)8
C)
9 D
)10
E
)11
48
4
(1)
?1
n
nn
n
21.
Wh
at is
th
e va
lue
of t
he
prod
uct
A)
[(p
–1)
! ]2B
)[(
p–
1)!
]3C
)(p
!)3
D)
(p!)
2E
)(p
!)3
– [(
p–
1)!]
3
12
3
1
()?
p k
pkk
13.
Wh
at is
the
valu
e of
the
prod
uct
A)
B)
C)
D)
E)
191
2917
629
168
2516
324
156
24
25
25
7(1
)?16
kk
17.
If f
: +
+,
and
g:
+
then
wh
at is
the
valu
e of
g(17
)?
A)
B)
C)
D)
E)
17
18!
218
18!
217
17!
218
17!
217
18!
32
2
()
()
()
(1)
x
k
fk
gx
fk
fk
1
x k
k
19.
Wh
at is
th
e re
sult
of
the
sum
A)
3136
B)
3442
C)
3558
D)
3726
E
)38
29
103
2
10
(25
41)
?k
kk
k
18.
Wh
at is
th
e va
lue
of t
he
prod
uct
A)
31024
–1
B)
31024
+1
C)
D)
E)
1024
31
4
1024
31
2
1024
31
2
9(2
)
0
(31)
?k
k
16.
Wh
at is
th
e va
lue
of t
he
sum
w
her
e
i=
ò–1
?
A)
2+
2ni
B)
2n+
1C
)n
+2i
D)
2n+
1 –
2n
iE
)2n
2+
1 –
2n
i
4
0
((1)
)n
k
k
ki
14.
If
then
wh
at i
s th
e su
m o
f th
e
coef
ficie
nts
of th
e te
rms
of P
(x)
with
eve
n po
wer
s?
A)
53·
14!
B)
–53
·13
!C
)–5
6·
13!
D)
56 ·
13!
E)
–56
·14!
14
2
()
()
k
Px
xk
20.
then
wh
at is
th
e
nu
mer
ical
val
ue
of a
– b
+ c
?
A)
8B
)14
C
)21
D
)29
E
)34
92
8
(52)
n k
kan
bnc
10.
Wh
at is
th
e va
lue
of t
he
sum
A)
188
B)
202
C)
249
D)
286
E)
304
12
2
?2
n
n
144
Appl
ied
Mat
hem
atic
s 11
24.
Wh
at is
th
e va
lue
of t
he
prod
uct
A)
1B
)1,
1 C
)1,
01
D)
1,11
E
)1,
101
110
0 1
(1)
(1)?
k
kk
30.
Wh
ich
on
e of
th
e fo
llow
ing
is e
qual
to
prod
uct
wh
ere
A)
B)
C)
D)
E)
1
nk
k
k1
nk
k
n
1
!n k
k1
(1)
nk
k
k1
nn
k
k
!(
, )
?(
)!n
Pn
rn
r1
(,
)n r
Pn
r
28.
Wh
at is
th
e va
lue
of t
he
sum
1 +
(1
+ 2
) +
(1
+ 2
+ 2
2 ) +
...
+ (
1 +
2 +
22
+ ..
. + 2
n–
1 )?
A)
2nB
)2n
–n
C)
2n+
1–
n
D)
2n+
1 –
n–
2E
)n
· 2n
27.
If
then
wh
at is
the
valu
e of
the
sum
in
ter
ms
of m
?
A)
B)
C)
D)
E)
224
54
m19
25
2m
144
32
m28
83
2m
176
32
m
211
0
46
52
2k
kk
k
11
11
11
...2
34
1112
m
26.
then
wh
at is
th
e va
lue
of f
(5)?
A)
27B
)54
C
)81
D
)16
2 E
)24
3
1
()
3n
n
x
fx
29.
If n
cor
ds o
f a
circ
le a
re d
raw
n, t
hen
let
t nbe
th
e
max
imu
m n
um
ber
of n
on-o
verl
apin
g re
gion
s th
at
can
be
form
ed w
ith
in t
he
circ
le il
lust
rate
d be
low
.
Wh
at is
th
e va
lue
of t
10?
A)
52B
)54
C
)56
D
)58
E
)60
1 c
hord
t 2 =
2 r
egi
on
s
2 c
hord
s
t 2 =
4 r
egi
on
s
3 c
hord
s
t 3 =
7 r
egi
on
s
25.
If t
he
sum
1 +
2 +
3+
...
+ n
= M
then
wh
at i
s
the
valu
e of
th
e su
m
n+
(n
+ 3
) +
(n
+ 6
) +
(n
+ 9
) +
...
+ 4
nin
term
s of
M?
A)
3MB
)4M
C
)5M
D
)6M
E
)7M
23.
Pen
tago
nal
nu
mbe
rs c
an b
e re
pres
ente
d by
dot
s
that
are
arr
ange
d in
th
e sh
ape
of a
pen
tago
n,
as
show
n b
elow
. T
he
firs
t fo
ur
pen
tago
nal
nu
mbe
rs
are
give
n. W
hat
is t
he
ten
th p
enta
gon
al n
um
ber?
A)
128
B)
130
C)
145
D)
164
E)
182
t 1 =
1t 2
= 5
t 3 =
12
t 4 =
22
22.
Wh
at is
th
e va
lue
of t
he
sum
A)
B)
C)
D)
E)
35 9
18 8
34 8
34 8
33 9
25
0
8?
8k
k
