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