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Mathematics
Session
Binomial Theorem Session 1
Session Objectives
Session Objective
1. Binomial theorem for positive integral index
2. Binomial coefficients — Pascal’s triangle
3. Special cases
(i) General term
(ii) Middle term
(iii) Greatest coefficient
(iv) Coefficient of xp
(v) Term dependent of x
(vi) Greatest term
Binomial Theorem for positive integral index
For positive integer n
n n n 0 n n–1 1 n n–2 2 n 1 n–1 n 0 n0 1 2 n–1 na b c a b c a b c a b ... c a b c a b n
n n–r rr
r 0
c a b
where
n nr n–r
n! n!c c for 0 r n
r! n – r ! n – r !r!
are called binomial coefficients.
nr
n n – 1 ... n – r 1C ,
1.2.3...r
numerator contains r factors
Any expression containing two terms only is called binomial expression eg. a+b, 1 + ab etc
Binomial theorem
10 10 107 10–7 3
10! 10.9.8C 120 C C
7! 3! 3.2.1
Pascal’s Triangle
0C0
1C0 C1
1
C2
2
C3
3
C4
4
C5
5
C2
1
C3
1
C4
1
C5
1C
52
C5
3C
54
C4
3
C3
2
C4
2
2C0
3C0
4C05C0
0a b 1
1a b 1a 1b
2 2 2a b 1a 2ab 1b
3 3 2 2 3a b 1a 3a b 3ab 1b
4 4 3 2 2 3 4a b 1a 4a b 6a b 4ab 1b
5 5 4 3 2 2 3 4 5a b 1a 5a b 10a b 10a b 5ab 1b
n n n 1r–1 r rc c c
3
4
5
6
10
1
1
1
1
1
1
1
2
3
4
10 5
1 1
1 1
Observations from binomial theorem
1. (a+b)n has n+1 terms as 0 r n
2. Sum of indeces of a and b of each term in above expansion is n
3. Coefficients of terms equidistant from beginning and end is same as ncr = ncn-r
n n n 0 n n–1 1 n n–2 2 n 1 n–1 n 0 n0 1 2 n–1 na b c a b c a b c a b ... c a b c a b
Special cases of binomial theorem
n nn n n n–1 n n–2 2 n n0 1 2 nx – y c x – c x y c x y ... –1 c y
n
r n n–r rr
r 0
–1 c x y
n
n n n n 2 n n n r0 1 2 n r
r 0
1 x c c x c x ... c x c x
in ascending powers of x
n n n n n–1 n0 1 n1 x c x c x ... c
nn n–r
rr 0
c x
nx 1
in descending powers of x
Question
Illustrative Example
Expand (x + y)4+(x - y)4 and hence
find the value of 4 42 1 2 1
Solution :
4 4 4 0 4 3 1 4 2 2 4 1 3 4 0 40 1 2 3 4x y C x y C x y C x y C x y C x y
4 3 2 2 3 4x 4x y 6x y 4xy y
Similarly 4 4 3 2 2 3 4x y x 4x y 6x y 4xy y
4 4 4 2 2 4x y x y 2 x 6x y y
4 4 4 2 2 4Hence 2 1 2 1 2 2 6 2 1 1
=34
General term of (a + b)n
n n r rr 1 rT c a b ,r 0,1,2,....,n
n n 01 0r 0, First Term T c a b
n n 1 12 1r 1, Second Term T c a b
n n r 1 r 1r r 1T c a b ,r 1,2,3,....,n
1 2 3 4 5 n n 1
r 0 1 2 3 4 n 1 n
T T T T T T T
kth term from end is (n-k+2)th term from beginning
n+1 terms
Question
Illustrative Example
Find the 6th term in the
expansion of
and its 4th term from the end.
95
2x 4x5
Solution :9 r r
9r 1 r
4x 5T C
5 2x
4 5 4 59
6 5 1 5 4 54x 5 9! 4 5
T T C5 2x 4!5! 5 2 x
39.8.7.6 2 .5
4.3.2.1 x 5040
x
Illustrative Example
Find the 6th term in the
expansion of
and its 4th term from the end.
95
2x 4x5
Solution :9 r r
9r 1 r
4x 5T C
5 2x
4th term from end = 9-4+2 = 7th term from beginning i.e. T7
3 6 3 69
7 6 1 6 3 6 34x 5 9! 4 5
T T C5 2x 3!6! 5 2 x
3
39.8.7 5
3.2.1 x
310500
x
Middle term
CaseI: n is even, i.e. number of terms odd only one middle term
thn 2
term2
CaseII: n is odd, i.e. number of terms even, two middle terms
n nn 2 2
n 2 n n1
2 2 2
T T c a b
thn 1
term2
n 1 n 1n 2 2
n 1 n 1 n 11
2 2 2
T T c a b
thn 3
term2
n 1 n 1n 2 2
n 3 n 1 n 11
2 2 2
T T c a b
Middle term
= ?2n
1x
x
Greatest Coefficientn
rc , 0 r n
CaseI: n even
n1
2
nterm T is max i.e. for r
2Coefficient of middle
nn
2
C
CaseII: n odd
n 1 n 3
2 2
n 1 n 1term T or T is max i.e. for r or
2 2
Coefficient of middle
n nn 1 n 1
2 2
C or C
Question
Illustrative Example
Find the middle term(s) in the
expansion of and
hence find greatest coefficient in the expansion
73x3x
6
Solution :
Number of terms is 7 + 1 = 8 hence 2 middle terms, (7+1)/2 = 4th and (7+3)/2 = 5th
33 4 13
474 3 1 3 3
x 7! 3 xT T C 3x
6 4!3! 6
13
133
7.6.5 3x 105x
3.2.1 82
Illustrative Example
Find the middle term(s) in the expansion
of and hence find greatest
coefficient in the expansion
73x3x
6
Solution :
1515
47.6.5 x 35
x3.2.1 482 3
43 3 15
375 4 1 4 4
x 7! 3 xT T C 3x
6 3!4! 6
Hence Greatest coefficient is 7 7
4 37! 7.6.5
C or C or 353!4! 3.2.1
Coefficient of xp in the expansion of (f(x) + g(x))n
Algorithm
Step1: Write general term Tr+1
Step2: Simplify i.e. separate powers of x from coefficient and constants and equate final power of x to p
Step3: Find the value of r
Term independent of x in (f(x) + g(x))n
Algorithm
Step1: Write general term Tr+1
Step2: Simplify i.e. separate powers of x from coefficient and constants and equate final power of x to 0
Step3: Find the value of r
Question
Illustrative Example
Find the coefficient of x5 in the expansion
of and term independent of x10
23
13x
2x
Solution :
r10 r10 2
r 1 r 31
T C 3x2x
r
10 10 r 20 2r 3rr
1C 3 x
2
For coefficient of x5 , 20 - 5r = 5 r = 3
310 10 3 5
3 1 31
T C 3 x2
Coefficient of x5 = -32805
Solution Cont.
r10 r10 2
r 1 r 31
T C 3x2x
r10 10 r 20 2r 3r
r1
C 3 x2
For term independent of x i.e. coefficient of x0 , 20 - 5r = 0 r = 4
410 10 4
4 1 41
T C 32
Term independent of x 76545
8
Greatest term in the expansion
Algorithm
Step1: Find the general term Tr+1
Step2: Solve for r 1r+1
r
TT
Step3: Solve for r 1r+1
r+2
TT
Step4: Now find the common values of r obtained in step 2 and step3
Question
Illustrative Example
Find numerically the greatest term(s) in the expansion of (1+4x)8, when x = 1/3
Solution :
r8r 1 rT C 4x
r8rr 1
r 18r r 1
C 4xT
T C 4x
8!
4x8 r !r! 9 r 4x
8! r9 r ! r 1 !
36 4r
3r
1
36r
7
Solution Cont.
36r
7
r8rr 1
r 18r 2 r 1
C 4xT
T C 4x
8!8 r !r! r 18! 8 r 4x4x
7 r ! r 1 !
3r 3
32 4r
1
29r
7
29 36r
7 7 r = 5 i.e. 6th term
5 58
6 5 1 54 4
T T C 563 3
Class Test
Class Exercise 1
Find the term independent of
x in the expansion of 81 –1
3 51x x .
2
Solution :
8 r r1 18 3 5
r 1 r1
T C x . x2
8 r r8 r8 3 5
r1
C x2
40 5r 3r 40 8r8 r 8 r8 815 15
r r1 1
C x C x2 2
For the term to be independent of x40 8r
0 r 515
Hence sixth term is independent of x and is given by3
86 5
1 8! 1T C . 7
2 5! 3! 8
6T 7
Class Exercise 2
Solution :
Find (i) the coefficient of x9 (ii) the term independent of x, in the expansion
of 9
2 1x
3x
r9 r9 2
r 1 r1
T C x3x
r9 18 2r r
r1
C x3
r9 18 3r
r1
C x3
i) For Coefficient of x9 , 18-3r = 9 r = 3
3 39 9 9 9
4 r1 9! 1 28
T C x x x3 3! 6! 3 9
hence coefficient of x9 is -28/9
ii) Term independent of x or coefficient of x0
, 18 – 3r = 0 r = 6 6 6
97 6
1 9! 1 28T C
3 6! 3! 3 243
Class Exercise 3
Solution :
n n nr r 1 r 2For 2 r n, C 2 C C
n 1 n 1 n 2 n 2r 1 r 1 r ra) C b) 2 C c) 2 C d) C
n n nr r 1 r 2C 2 C C
n n n nr 1 r r 2 r 1C C C C
Now as n n n 1m m 1 m 1C C C
n n n nr 1 r r 2 r 1C C C C
n 1 n 1 n 2r r 1 rC C C
Class Exercise 4
Solution :
If the sum of the coefficients in the expansion of (x+y)n is 4096, then prove that the greatest coefficient in the expansion is 924. What will be its middle term?
Sum of the coefficients is n 122 4096 2
n 12 i.e. odd number of terms
greatest coefficient will be of the middle term
126
12! 12.11.10.9.8.7C 924
6! 6! 6.5.4.3.2.1
Middle term = 12 6 6 6 66C x y 924 x y
Class Exercise 5
Solution :
If
then prove that
n2 2 2n0 1 2 2n1 x x a a x a x ... a x
n
0 2 4 2n3 1
a a a ... a2
n2 2 2n0 1 2 2n1 x x a a x a x ... a x ...(i)
Replace x by –x in above expansion we get
n2 2 2n0 1 2 2n1 x x a a x a x ... a x ...(ii)
Adding (i) and (ii) we get n n2 21– x x 1 x x
2 4 2n0 2 4 2n2 a a x a x ... a x
Solution Cont.
n n2 21– x x 1 x x
2 4 2n0 2 4 2n2 a a x a x ... a x
Put x = 1 in above, we get
n0 2 4 2n1 3 2 a a a ... a
n
0 2 4 2n3 1
a a a ... a2
Class Exercise 6
Solution :
Let n be a positive integer. If the coefficients of 2nd, 3rd, 4th terms in the expansion of (x+y)n are in AP, then find the value of n.
n n 2 n 32 1 3 2 4 3T C x, T C x , T C x n n n
1 2 3C , C , C are in AP
n n n2 1 32 C C C
2 n! n! n!2! n – 2 ! 1! n – 1 ! 3! n – 3 !
1 1 1
n – 2 ! n – 1 ! 3! n – 3 !
6 n – 1 n – 21
n – 2 ! 3! n – 1 !
26 n – 1 6 n – 3n 2 2x – 9n 14 0 n – 7 n – 2 0
n 2 or n 7 n 7
Class Exercise 7
Solution :
Show that
Hence show that the integral part of
is 197.
6 62 1 2 – 1 198.
62 1
6 6LHS 2 1 2 – 1
6 5 4 3 26 6 6 61 2 3 42 C 2 C 2 C 2 C 2
6 56 6 6 65 6 0 1C 2 C C 2 – C 2
4 3 2 16 6 6 6 62 3 4 5 6C 2 C 2 C 2 C 2 C
6 4 26 6 62 4 62 2 C 2 C 2 C 6 6
2 1 2 1
Solution Cont.
6 4 26 6 62 4 62 2 C 2 C 2 C
3 26! 6!2 2 .2 .2 1
2! 4! 4! 2!
= 2 (8 + 15.4 + 15.2 + 1) = 198 = RHS
Let 62 1 I f where
I = Integral part of 62 1 and f = fraction part of
62 1 0 f 1 i.e.
60 2 – 1 1 0 2 – 1 1
Solution Cont.
f f´ 198 – I Integer
Now as '0 f 1 and 0 f 1
let 62 – 1 f´
6 62 1 2 – 1 I f f´ 198
'0 f 1 and 0 f 1
f f´ is an integer lying between 0 and 2
f f´ 1 I 198 – f f´ 198 – 1́ 197
Integer part of is 197. 62 1
0 f f ' 2
Class Exercise 8
Solution :
Find the value of greatest term
in the expansion of 20
13 1
3
Consider 20
11
3
Let Tr+1 be the greatest term r 1 r r 1 r 2T T and T T
r 1 rT T 3 r–1
20 20r r–1
1 1C C
3 3
20! 1 20!
.r! 20 – r ! r – 1 ! 21– r !3
20! 1 20!
.r! 20 – r ! r – 1 ! 21– r !3
Solution Cont.
1 1 1.
r 21– r3 21– r 3 r
21 3 – 121 21 3 – 21r
2 23 1
r 1 r 2T T r r 1
20 20r r 1
1 1C C
3 3
20! 20! 1
.r! 20 – r ! r 1 ! 19 – r ! 3
1 1 1.
20 – r r 1 3
3 r 3 20 – r 20 – 3 3 – 120 – 3 21 3 – 23
r3 – 1 23 1
Solution Cont.
21 3 – 23 21 3 – 21r
2 2
6.686 r 7.686
r = 7 is the only integer value lying in this interval
720
8 71
T 3 C3
is the greatest term.
Class Exercise 9If O be the sum of odd terms and E that of even terms in the expansion of (x + b)n prove that
n2 2 2 2O – E x – b
2n 2n4OE x b – x – b
2n 2n2 22 O E x b x – b
i)
ii)
iii)
nx b n n 0 n n 1 1 n 1 n 1 n 0 n0 1 n 1 nC x b C x b ... C x b C x b
Solution :
n n 0 n n–2 20 2O C x b C x b ...
n n–1 1 n n–3 31 3E C x b C x b ...
Solution Cont.
n n 0 n n–2 20 2O C x b C x b ...
n n–1 1 n n–3 31 3E C x b C x b ...
nO E x b O - E = (x-b)n
n n2 2O – E O E O – E x b x – b n2 2x b
4 OE = 2 2 2n 2nO E – O – E x b – x – b
2 22 22 O E O E O – E 2n 2nx b x – b
Class Exercise 10
Solution :
In the expansion of (1+x)n the binomial coefficients of three consecutive terms are respectively 220, 495 and 792, find the value of n.
Let the terms be r r 1 r 2T , T , T
n r–1 nr r–1 r–1T C x C 220
n r nr 1 r rT C x C 495
n r 1 nr 2 r 1 r 1T C x C 792
n
r r–1n
r 1 r
r! n – r !T C n! 220.
T r – 1 ! n – r 1 ! n! 495C
Solution Cont.
n
r r–1n
r 1 r
r! n – r !T C n! 220.
T r – 1 ! n – r 1 ! n! 495C
r 220 4n – r 1 495 9
9r 4n – 4r 4
4n 4r ...(i)
13
Similarly r 1
r 2
T r 1 495 5
T n – r 792 8
5n – 88r 8 5n – 5r r
13 ...(ii)
Solution Cont.
4n 4r ...(i)
13
5n 8r ... ii
13
From (i) and (ii)
4n 4 5n – 813 13
4n 4 5n – 8
n = 12
Thank you