Upload
ahadh12345
View
218
Download
0
Embed Size (px)
Citation preview
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
1/23
1. (a)
9
4
12
+ x
= 29+ 9 2
8
x4
1
+ 2
89
(27) 16
2x
+ 6
789
26 64
3x
M1 B1
(M1 for descending powers of 2 and ascending powers ofx; B1 for
coefficients 1 9 36 84 in an! for" as a#o$e)
= %12 + %76x + 288x2
+ 84x3
&1 &1 4
(#) x= 1'1
gi$es
(2'2%)9
= %12 + %76 + 288 + ''84 M1
= %72%64 &1 2[6]
2. (a) ar= 9
ar4= 112% M1
i$iding gi$es 8
1
9
12%13 ==rM1
*o r= 21
&1 3
(#) sing ar= 9 a= 21
9
= 18 M1 &1 2
(c) S
= 211
181 = r
a
= 36 M1 &1 2
[7]
3. (a)
32
,3
)2)(1(
,2
)1(k
nnnk
nn =
One coefficient (no
r
n
) M1A correct equation, no cancelling A1
eg 3k2= (n2)k3
Cancel at least n (n 1) M1
3 = (n- 2)k (*) &1 cso 4
(#) A= nk= 4 B1
3 = 4 - 2k M1
*o k = 21
and n= 8 &1 &1 4[8]
Holland Park School 1
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
2/23
4. (a)96'
1
12''
1=
=
rra
M1 &1
96' (1 r) = 12'' r= 41
(.) &1
(#) /9= 12'' (- '2%)
8(or /
1') M1
ifference = /9/
1'= ''1831'%0('''4%7760) M1
= ''23 (or ''23) &1
(c) * n =
( )(( )2%'1
2%'112''
n
M1 &1
(d) *ince nis odd ('2%)nis negati$e M1
so *n= 96' (1 + '2%n
) (.) &1[10]
5. (1 +px)n1 + npx + 2
)1(22
xpnn
+ 0 B1 B1
o"paring coefficients np= 18 2)1( nn
= 36 M1 &1
*o$ing n(n- 1) = 72 to gi$e n= 9;p= 2 M1 &1; &1 ft[7]
6. (2 -px)6= 2
6+
1
6
2%(-px) +
2
6
24(-px)
2M1
r
n
oa!
Coeff. of x or x2
= 64 + 6 2%(-px); + 1% 2
4(-px)
2&1; &1
o
rn
1% 16p2= 13% p2= 16
9
or p = 43
(on!) M1 &1
-632p=A A = -144 M1 &1 ft (t5eirp( ')) 7Con!one lost or extra "" signs for M #arks $ut A #arks #ust
$e correct.
%inal A1 ft is for 1&2x (t'eir p )[7]
Holland Park School 2
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
3/23
7. (a) 1 + nax + 2
)1( nn
(ax)2+ 6
)2)(1( nnn
(ax)3+ 0 B1B1 2
accept 2*, +*
(#) na= 8 2
)1( nn
a2= 3' M1
$ot'
2
64
2
)1(
n
nn
= 3'
( )2
1 288 aaa
= 3' M1
eit'er
n= 16 a= 2
1
&1 &1 4
(c)
3
2
1
6
141%16
= 7' M1 &1 2[8]
8. (a) (S=) a+ ar+ 0 + arn-1
B1
S =- not require!. A!!ition require!.
(rS= ) ar+ ar2+ 0 + ar
nM1
rS =- not require! (M Multipl/ $/ r)
S(1 - r) = a(1 - rn) S= r
ra n
1
)1(
M1 &1 4
(M Su$tract an! factorise eac' si!e) (0)
(#) r= '9 B1
S2'
= 9'1
)9'1(1' 2'
= 878 M1 &1 3
(c) *" to infinit! = 9'1
1'
1 =
ra
= 1'' M1 &1ft 2
(ft onl/ for r 1)
Holland Park School 3
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
4/23
(d) r
r
r
a
=
11 = 1' (t a= rin t5e for"a fro" (c) and eate to 1') M1
r= 1'(1 - r) r= 0 11
1'
(or e:act ei$aent) M1 &1 3[12]
9. (a) (x3)12
; 0 +
2
1)()(
2
12
2
1)()(
1
122
1'3113 +
+
x
xx
x
B1; M1
%or M1, nee!s $ino#ial coefficients,nC
rfor# O3 at least as
far as s'o4n5
orrect $aes fornC
rs 12 66 22' sed ("a! #e i"pied) B1
(x3)12
+ 12(x3)11 2
1)(22'2
1)(662
13
93
2
1'3 ++ xxxxx
=
24283236
2
%%
2
336 xxxx +
&2(1') %
(#) /er" in$o$ing
9
33
2
1)()(
xx
; M1
coeff =
9
2
1)(
123
1'1112
&1
=128
%%
(or -'429687%) &1 3[8]
10. (a) ar3= 12 ar
4= -8 r= 0 3
2 (or e:act ei$aent) M1 &1 2
(#) sing rwit5 ar3= 12 or ar
4= -8 to find a= 0 M1
a =-4' 1
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
5/23
11. (a)
2
1'2
1'2
%2
3
2
2
34%
%
+
+
++=
+
xk
xk
xkk
xk
M1 &1 2
eed not #e si"pified &ccept%
1
%
2
%
3
(#) 1'k3
2
2
x
= %4'x2wit5 or wit5otx
2M1
>eading to k= 6 (.) cso &1 2
?r s#stitting k= 6 into%
2k
3
2
2
x
and si"pif!ing to %4'x2
(c) oefficient ofx3is 1' 6
2
32
1
= 4% M1 &1 2[6]
12. (a) &pp!ing correct for"a @32% = 12' +%(n - 1)A M1
*o$ing to gi$e n= 42 (.) (or $erif!ing in correct eation) &1 2
(#) sing for"a for s" of & * = 2
42
24' + %(42 - 1)
or se 2 n a l+
M1&1
= 934% &1 3
(c) Cecognising D wit5 r= '98 M1
Eae ( in F ) = 72'' ( '98)24
M1
= 4434 ( on! t5is $ae) &1 3[8]
13. (3 + 2x)%= (3
%) +
1
%
34 (2x) +
2
%
33(2x)
2+ 0 M1
= 243+81'x+1'8'x2 B1 &1 &1 4
[4]
14. (a) ar72 ar3= %832 r2= 27
832%
(= '81) M1
r= '9 &1 2
(#) a=)(
27
a = 8 M1 &1 2
Holland Park School 5
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
6/23
(c) s%'
= 9'1
))9'(1(8%'
M1
= 79%88 (3!p) &1 cao 2
(d) s
= 9'1
8
(= 8') M1
s
-s%'
= 8' - (c) = '412 &1 ft 2
[8]
15. (a) 1 + 12px +66p2x
2B1 B1 2
accept an/ correct equi6alent
(#) 12p= -q 66p2= 11q M1
%or#ing 2 equations $/ co#paring coefficients
*o$ing forpor q M1
p= -2 q= 24 &1&1 4[6]
16. (a) Grites down #ino"ia e:pansion p to and incding ter" inx3
aownC
rnotation 1 + nax+ n(n- 1) 6
)2)(1(
2
22 +
nnnxa
a3x
3
(con!one errors in po4ers of a)*tates na=1% B1
ts 6
)2)(1(
2
)1( 32 annnann =
dM1
(con!one errors in po4ers of a )
3 = (n2 )a
*o$es si"taneos eations in nand ato o#tain
a=6 and n= 2% M1 &1 &1 6
n.$. 7ust 4rites a = 8, an! n = 2.9 follo4ing no
4orking or follo4ing errors allo4 t'e last M1 A1 A15
(#) oefficient ofx3= 2% 1% '% 6
3H 6 = 67% B1 1
(or equals coefficient of x2= 2.9 : 1.9 : 8
2; 2 = 8
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
7/23
17. (a) (S=) a+ ar+ 0 + arn-1
B1
S =- not require!.
A!!ition require!
(rS=) ar+ ar2+ 0 + ar
nM1
rS =- not require!
(M Multipl/ $/ r)
S(1 - r) = a(1 - rn) S= r
ra n
1)-1(
M1 &1cso 4
(M Su$tract an! factorise) (0)
1 At least t'e + ter#s s'o4n a$o6e, an! no extra
ter#s.
A1 >equires a co#pletel/ correct solution.
Alternati6e for t'e 2 M #arks
M1 Multipl/ nu#erator an! !eno#inator $/ 1 r.
M1 Multipl/ out nu#erator con6incingl/, an!factorise.
(#) arn-1
= 3%''' 1'43= 39 4'' M1 &1 2
(M Correct a an! r, 4it' n = +, ? or 9).
M1 can also $e score! $/ a /ear $/ /ear- #et'o!.
Ans4er onl/ +& ? scores full #arks, +& +
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
8/23
18. (2 +x)6= 64 B1
(6 2%x) +
22
%6 24
x
+192x +24'x2
M1&1&1 4
B'e ter#s can $e "liste! rat'er t'an a!!e!.
M1 >equires correct structure "$ino#ial coefficients (per'aps
fro# ascals triangle), increasing po4ers of one ter#,
!ecreasing po4ers of t'e ot'er ter# (t'is #a/ $e 1 if factor 2
'as $een taken out). Allo4 "slips.
1
6
an!
2
6
or equi6alent are accepta$le,
or e6en
1
6
an!
2
6
Decreasing po4ers of x
Can score onl/ t'e M #ark.
8?(1 EFF), e6en if all ter#s in t'e $racket are correct, scores
#ax. 1M1AA.
[4
19. (a) ar = 4
2%1
= ra
(/5ese can #e seen esew5ere) B1 B1
a = 2%(1 r) 2%r(1 r) = 4 M Ii"inate a M1
2%r22%r + 4 = ' &1cso 4B'e M #ark is not !epen!ent,
$ut $ot' expressions #ust contain $ot' a an! r.
(#) (%r)(%r4) = ' r= %1
or %
4
M1&1 2
Special case
One correct r 6alue gi6en, 4it' no #et'o!
(or per'aps trial an! error) 1 .
(c) r= a= 2' or % M1&1 2M1 Su$stitute one r 6alue $ack to fin! a 6alue of a.
(d) r
raS
n
n
=1
)1(
#t
2%1
= ra
so Sn= 2%( rn) B1 1
Sufficient 'ere to 6erif/ 4it' Gust one pair of 6alues of a an! r.
(e) 2%(1 '8n) 24 and proceed to n= (or or J) wit5 no nsond age#ra M1
Holland Park School 8
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
9/23
=> )42%14(
8'og
'4'ogn
n= 1% &1 2
Accept =- rat'er t'an inequalities t'roug'out, an! also allo4
t'e 4rong inequalit/ to $e use! at an/ stage.
M1 requires use of t'eir larger 6alue of r.
A correct ans4er 4it' no 4orking scores $ot' #arks.%or trial an! error- #et'o!s, to score M1, a 6alue of n
$et4een 12 an! 1H (inclusi6e) #ust $e trie!.
[11]
20. (a)
&ppropriate figre M1
1
1
sin i i
i i
r r
r r +
+
=
+(e:p for sin) M1 &1
ri+ r
i+1)sin= r
i- r
i+1
+i
i
r
r 1
M1
1 sin
ration of radii . (=r)1 sin
= + &1 cso %
(#) /ota area2 2 2
2 3 > r r = + + +
2 2 4(1 ) (correct KrK)> r r= + + + B12
2
2 2
1
1 1 sin1
1 sin
>>
r
= = + M1
2 2 2 2
2 2
(1 sin ) (1 sin )
4sin(1 sin ) (1 sin )
> >
+ += =
+ &1 3
Holland Park School 9
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
10/23
(c) Ceired area = 2 &rea OA+ &rea "aLor sectorAO
&rea fond in (#) M1
&rea OA= 21
>(>cot ) B1
ange of "aLor sector to B 22
OA = = +M1
&rea sectorAO= 21
>2(+ 2) &1
Ceired area =>2(cot +
+++
sin
sinsin21
42
2
!
&1 cso %
=>2(+ cot-
sin4
eccos4
)
(d)
+=
cos
4cotcosec
4cosec1 22>
!
!s
M1 &1
2 2 cos( cot cos )
4 sin 4>
= +
2 2 2( cot cos (cosec 1))4
>
= + M1
2 2
cot ( cos 1)4>
= &1 (cso) 42(se of cot cosec 1) oe=
Holland Park School 10
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
11/23
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
12/23
(a) 2nd
B1 for
2
9
(px)2or #etter ondone OP not O+P
(#) 1stM1 for a inear eation forp
2nd
M1 for eit5er printed e:pression foow t5rog5 t5eirp
B 1 + 9px2+ 36px
2eading top= 4 q= 144 scores B1B' M1 &1M1&' ie 4
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
13/23
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
14/23
4'x2(1
st&1) &1
-8'x3(2
nd&1) &1
&ow co""as #etween ter"s /er"s "a! #e isted rat5er t5an
added
&ow Treco$er!U fro" in$isi#e #racets so
32234% 213
%21
2
%21
1
%1 xxx
+
+
+
= 1 - 1'x + 4'x2- 8'x
3+
gains f "ars
1 + % (2x) +)2(
,3
34%)2(
,2
4% 32 +
+
xx
= 1 + 1'x + 4'x2+ 8'x
3+
gains B'M1&1&'
Misread first 4 ter"s descending ter"s if correct wod score
B' M1 1st&1 one of 4'x
2and -8'x
3correct; 2
nd&1 #ot5 4'x
2and
-8'x3correct
>ong "tipication
(1 - 2x)2= 1 - 4x + 4x
2 (1 - 2x)
3= 1 - 6x + 12x
2- 8x
3
(1 - 2x)4= 1 - 8x + 24x
2- 32x
3+ 16x
4
(1 - 2x)%= 1 - 1'x + 4'x
2+ 8'x
3+ 0
1 - 1'x
1 - 1'x "st #e seen in t5is si"pified for" in (a) B1
&tte"pt repeated "tipication p to and incding (1 - 2x)%
M1
4'x2(1
st&1) &1
- 8'x3(2
nd&1) &1
Misread first 4 ter"s descending ter"s if correct wod score
B' M1 1st&1 one of 4'x
2and -8'x
3correct;
2nd
&1 #ot5 4'x2and -8'x
3correct
Holland Park School 14
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
15/23
(#) (1 +x)(1 - 2x)%= (1 +x)(1 - 1'x + 0)
= 1 +x- 1'x + 0 M1
1 - 9x (.) &1 2
se t5eir (a) and atte"pt to "tip! ot; ter"s (w5et5er correct
or incorrect) inx2or 5ig5er can #e ignored
f t5eir (a) is correct an atte"pt to "tip! ot can #e i"pied fro"t5e correct answer so (1 +x)(1 - 1'x) = 1 - 9x wi gain M1 &1
f t5eir (a) is correct t5e 2nd #racet "st contain at east (1 - 1'x)
and an atte"pt to "tip! ot for t5e M "ar &n atte"pt to "tip! ot is an atte"pt at
2 ot of t5e 3 ree$ant ter"s (B t5e 2 ter"s in x1"a! #e
co"#ined - #t t5is wi sti cont as 2 ter"s)
f t5eir (a) is incorrect t5eir 2nd
#racet "st contain a t5e ter"s
inx'andx
1fro" t5eir (a)
& an atte"pt to "tip! a ter"s t5at prodce ter"s in x'andx
1
B (1 +x)(1 - 2x)%= (1 +x)(1 - 2x) @w5ere 1 - 2x + 0 is ?/
t5e candidateUs answer to (a)A
= 1 -xie candidate 5as ignored t5e power of % M'
B /5e candidate "a! start again wit5 t5e #ino"ia e:pansion for
(1 - 2x)%in (#) f correct (on! needs 1 - 1'x) "a! gain M1 &1
e$en if candidate did not gain B1 in part (a) M1
B &nswer gi$en in estion &1
I:a"pe
&nswer in (a) is = 1+1'x + 4'x2- 8'x
3+
(1 +x)(1 + 1'x) = 1 + 1'x +x M1= 1 + 11x &'
[6]
Holland Park School 15
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
16/23
25. (a) Sn= a + ar + 0 + ar
n- 1B1
rSn= ar + ar
2+ 0 + ar
nM1
(1 - r)*n= a(1 - r
n) dM1
Sn= r
ra n
1
)1(
(.) &1cso 4
Snnot reired /5e foowing "st #e seen at east one + sign
a arn- 1
and one ot5er inter"ediate ter" o e:tra ter"s (sa! arn) B1
Mtip! #! r; rSnnot reired
&t east 2 of t5eir ter"s on CV* correct! "tipied #! r M1
*#tract #ot5 sides >V* "st #e Q(1 - r)Sn CV* "st #e
in t5e for" Qa(1 - rpn + q
)
?n! award t5is "ar if t5e ine for Sn= 0 or t5e ine for rS
n= 0
contains a ter" of t5e for" arcn E !
Met5od "ar so "a! contain a sip #t not awarded if ast ter"
of t5eir Sn= ast ter" of t5eir rS
n dM1
o"petion cso
B &nswer gi$en in estion &1cso
Snnot reired /5e foowing "st #e seen at east one + sign
a arn- 1
and one ot5er inter"ediate ter" o e:tra ter"s (sa! arn) B1
?n CV* "tip! #! r
r
1
1
?r Mtip! >V* and CV* #! (1 - r) M1
Mtip! #! (1 - r) con$incing! (CV*) and tae ot factor of a
Met5od "ar so "a! contain a sip dM1
o"petion cso B &nswer gi$en in estion &1cso
(#) a= 2'' r= 2 n= 1' S1'
= 21
)21(2''1'
M1 &1
= 2'46'' &1 3
*#stitte r = 2 wit5 a = 1'' or 2'' and n = 9 or 1' into for"a for Sn. M1
21
)21(2''1'
or ei$aent &1
2'46'' &1
Holland Park School 16
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
17/23
&ternati$e "et5od adding 1' ter"s
(i) &nswer on! f "ars (M1 &1 &1)
(ii) 2'' + 4'' + 8'' + 0 + 1'24'' = 2'46'' or
1''(2 + 4 + 8 + 0 + 1'24) = 2'46''
M1 for two correct ter"s (as a#o$e oe) and an indication t5at t5e
s" is needed (eg + sign or t5e word s") M1
1'24'' oe as fina ter" an #e i"pied #! a correct fina answer &1
2'46'' &1
(c) 3
1
6
%== ra
B1
S
= 3
1
1
6
%
1
=
Sr
a
M1
4
%=
oe &1 3
B SW
= r
a
1 is in t5e for"ae #oo
r =3
1
seen or i"pied an!w5ere B1
*#stitte a = 6
%
and t5eir r into r
a
1 sa res a#ot oting for"a M1
4
%
oe &1
(d) -1 J rJ 1 (or rJ 1) B1 1
B SI
= r
a
1 for X r J J1 is in t5e for"ae #oo
-1 J r J 1 or X r X J 1 n words or s!"#os
/ae s!"#os if words and s!"#os are contradictor! Mst #e J not Y B1[11]
Holland Park School 17
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
18/23
26. (a) 1 + 6kx @&ow nsi"pified $ersions eg 16+ 6(1
%)kx
6C
'+
6C
1kxA B1
32 )(23
4%6)(
2
%6kxkx
+
+@*ee #eow for accepta#e $ersionsA M1&1 3
B /V* II ?/ BI *M>RI R?C /VI &1 (isw is appied)
/5e ter"s can #e TistedU rat5er t5an added
M1 Ceires correct strctre T#ino"ia coefficientsU
(per5aps fro" ascaUs triange) increasing powers ofx
&ow a TsipU or TsipsU sc5 as
3232
3232
23
4%6
2
%6
23
34%
2
4%
)(23
%6)(
2
%6
23
4%6
2
%6
xxkxkx
kxkxkxkx
+
+
+
+
+
+
+
=
Bt 1% + k2x
2+ 2' + k
3x
3or si"iar is M'
Bot5x2andx
3ter"s "st #e seen
3
6
and2
6
or ei$aent sc5 as6C
2and
6C
3are
accepta#e and e$en
3
6and
2
6
are accepta#e for t5e
"et5od "ar
&1 &n! correct (possi#! nsi"pified) $ersion of t5ese 2 ter"s
3
6and
2
6
or ei$aent sc5 as6C
2and
6C
3are accepta#e
escending powers ofx
an score t5e M "ar if t5e reired first 4 ter"s are not seen
Mtip!ing ot (1 + kx)(1 + kx)(1 + kx)(1 + kx)(1 + kx)(1 + kx)
M1 & f atte"pt to "tip! ot (power 6)
B1 and &1 as on t5e "ain sc5e"e
(#) 6k= 1%k2
k= %
2
(or ei$ fraction or '4) (gnore k= ' if seen) M1&1cso 2
M Iating t5e coefficients ofxandx2
(e$en if tri$iaeg 6k= 1%k)
&ow t5is "ar aso for t5e T"isreadU eating t5e
coefficients ofx2andx
3
&n eation is kaone is reired for t5is M "ar at5og5
condone 6kx= 1%k2x
2(6k= 1%k2) k= %
2
Holland Park School 18
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
19/23
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
20/23
(d) Sn= '911
)'911('''%'
1
)1(1'
=
r
ra n
M1&1
F76' ''' (Mst #e t5is answer nearest F1'''') &1 3
M1 se of t5e correct for"a wit5 a= %'''' %''' or %''''' and
n= 9 1' 11 or 1%
M1 can aso #e scored #! a O!ear #! !earP "et5od wit5
ter"s added
(&ow t5e M "ar if t5ere is e$idence of adding 9 1' 11 or 1% ter"s)
1st&1 is scored if 1' correct ter"s 5a$e #een added (aow
Onearest F1''P)
(%'''' %4%'' %94'% 647%1 7'%79 76931 838%% 914'2
99628 1'8%9%)
o woring s5own *pecia case 76' ''' scores 1 "ar scored as 1 ' '
(?t5er answers wit5 no woring score no "ars)[9]
28. (a) o"pete "et5od sing ter"s of for" ark to find r M1
@eg Dividing ar6= 8' #! ar
3= 1' to find r; r
6- r
3= 8 is M'A
r = 2 &1 2
M1 ondone errors in powers eg ar4= 1' and
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
21/23
29. (a)
321'
2
1
3
1'
2
1
2
1'
2
1
1
1'1
2
11
+
+
+=
+ xxxx
M1&1
= 1 + %x; 4
4%+
(or 112%)x2+ 1%x
3
(coeffs need to #e t5ese ie si"pified) &1; &1 4
@&ow &1 &' if tota! correct wit5 nsi"pified singefraction coefficients)
Ror M1 first &1 onsider nderined e:pression on!
M1 Ceires correct strctre for at east two of t5e t5ree ter"s
(i) Mst #e atte"pt at #ino"ia coefficients
@Be generos aow a notations eg1'
C2 e$en
2
1'
; aow OsipsPA
(ii) Mst 5a$e increasing powers ofx
(iii) Ma! #e isted need not #e added; t'is applies for all #arks
Rirst &1 Ceires all three correct ter"s #t need not #e si"pified
aow 11'
etc1'
C2etc and condone o"ission of #racets arond
powers of \x
*econd &1 onsider as B1 1 5x
(#) (1 + 2
1
'.'1)1'
= 1 + %(''1) + ( 4
4%
or 112%) (''1)2+ 1%(''1)
3M1&1ft
= 1 + ''% + '''112% + '''''1%= 1'%114 cao &1 3
Ror M1 *#stitting their (''1) into t5eir (a) rest
@'1 '''1 '2% ''2%'''2% accepta#e #t not '''% or 1''%A
Rirst &1 (ft) *#stittion of (0.01) into their 4 ter!ed e:pression in (a)
"n#$er $ith no $or%ing scores no !ar%# (cacator gi$es t5is answer)[7]
30. (a) (1 + ax)
1'
= 1 + 1'ax (ot nsi"pified $ersions) B132 )(
6
891')(
2
91'axax
+
+
I$idence fro" one of t5ese ter"s is sfficient M1
+ 4%(ax)2 + 12'(ax)
3or + 4%a
2x
2 + 12'a
3x
3&1 &1 4
Holland Park School 21
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
22/23
/5e ter"s can #e TistedU rat5er t5an added
M1 Ceires correct strctre T#ino"ia coefficientU (per5aps
fro" ascaUs triange) and t5e correct power ofx
(/5e M "ar can aso #e gi$en for an e:pansion in
descending powers ofx)
&ow TsipsU sc5 as
33232
23
789
2
91')(
23
91'
2
91'xaxaxax
Vowe$er 4% + a2x
2+ 12' + a
3x
3or si"iar is M'
3
1'and
2
1'
or ei$aent sc5 as1'
C2and
1'C
3
are accepta#e and e$en
3
1'and
2
1'
are accepta#e
for t5e "et5od "ar
1
st
&1 orrectx
2
ter"2
nd&1 orrectx
3ter" (/5ese "st #e si"pified)
f si"pification is not seen in (a) #t correct si"pified ter"s
are seen in (#) t5ese "ars can #e awarded Vowe$er if
wrong si"pification is seen in (a) t5is taes precedence
*pecia case
f (ax)2and (ax)
3are seen wit5in t5e woring #t t5en ost0
0 &1 &' can #e gi$en if 4%ax2and 12'ax
3are #ot5 ac5ie$ed
(#) 12'a3= 2 4%a
2a = 4
3
or ei$
7%'12'
9'eg
gnore a = ' if seen M1 &1 2
M Iating t5eir coefficent ofx3to twice t5eir coefficient
ofx20 or eating t5eir coefficent ofx
2to twice t5eir
coefficient ofx3 ( or coefficients can #e correct
coefficients rat5er t5an t5eir coefficients)
&ow t5is "ar e$en if t5e eation is tri$ia eg 12'a = 9'a
&n eation in a aone is reired for t5is M "ar
at5og5 condone eg 12'a3x
3= 9'a
2x
2(12'a3= 9'a2)
a= 4
3
Beware a = 4
3
foowing 12'a = 9'a w5ic5 is &'[6]
Holland Park School 22
8/13/2019 Core 2 - Ch 7 - 1 - Sequences and Series - Solutions
23/23
31. (a) B2'
=%
19
%
4
= ''72 (&ccept awrt) &ow % %
419
for M1 M1 &1 2
M Ceires se of t5e correct for"a arn-1
.
(#)2%
8'1
%=
=S
M1&1 2
M Ceires se of t5e correct for"a r
a
1
(a) and (#) orrect answer wit5ot woring scores #ot5 "ars
(c) 8'1
)8'1(%
k
249% (&ow wit5 = or J) M1
1 - '8k '998 (or ei$ see #eow) (&ow wit5 = or J) &1
k og '8 J og '''2 or k og'8
'''2 (&ow wit5 = or J) M1
k 8'og
''2'og
(.) &1cso 4
1stM /5e s" "a! 5a$e aread! #een T"anipatedU (per5aps
wrong!) #t t5is "ar can sti #e aowed
1st& & Tn"erica! correctU $ersion t5at 5as deat wit5
(1 - '8) deno"inator
eg 1 -
k
%
4
'998 %(1 - '8k) 499
2%(1 - '8k) 249% % - %('8
k) 499
n an! of t5ese %
4
instead of '8 is fineand condone
%
4k
if correct! treated ater
2nd
M ntrodcing ogs and sing aws of ogs correct!
(t5is "st incde deaing wit5 t5e power k so t5atp
k=k ogp).
2nd
& &n incorrect state"ent (incding eaities) at an!
stage in t5e woring oses t5is "ar (t5is is often
identifia#e at t5e stage k og '8 og '''2)
(*o a f! correct "et5od wit5 ineaities is reired)
(d) k =28 (Mst #e t5is integer $ae) ot k 27 or k J28 or k 28 B1 1[9]