aakash - binomial theorem and its application

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BINOMIAL EXPRESSION

An algebraic expression containing only two terms is called a binomial

expression. Such as a + x, 2x + 5y, y

x 12 , xba etc. all are binomal

expressions.

BINOMIAL THEOREM FOR POSITIVE INTEGRAL INDEX

If x, y R, then n N

(x + y)n = nC0xn + nC1x

n – 1y + nC2xn – 2y2 + .... + nCrx

n – ry r + .... + nCn yn

Here all nCr's are called binomial coefficient

)!(!!

rnrnCr

n

nC

CCn

nnn

1

0 1

nrCC rnn

rn 1;

Note : (i) n is the exponent or index of binomial

(ii) Total number of terms n + 1

(iii) The (r + 1)th term = Tr + 1 = nCrxn – ry r

(iv) Sum of exponents of x and y in any term = n

(v) Powers of x go on decreasing by 1 and powers of y go onincreasing by 1 in subsequent terms

(vi) In any term suffix of binomial coefficient is power of y

(a) Replacing y with –y in the expansion of (x + y)n, we get

(x – y)n = nC0xn – nC1x

n – 1y + nC2xn – 2y2 – ..... + (–1)n nCn y n

(b) Replacing x by 1 and y with x, we get

(1 + x)n = nC0 + nC1x + nC2x2 + ......... + nCnx

n

7Binomial Theorem andits Application

C H AP T E RAIEEE SyllabusBinomial theorem for a positive integral index, General term and middleterm, Properties of Binomial coefficients and Simple applications

THIS CHAPTERINCLUDES : Binomial expression,

Binomial thorem forpositive integralindex

General term in theexpansion of (x + y)n

Middle term in theexpansion of (x + y)n

pth term from theend in expansion of(x + y)n

Numerically greatestterm in theexpansion of (1 + x)n

Properties ofbinomial coefficients

Some relationinvolving binomialcoefficients

Some usefulexpansions.

Solved examples


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Binomial Theorem and its Application AIEEE/State CETs

(c) (1 – x)n = nC0 – nC1x + nC2x2 – .......... + (–1)n nCnx

n

(d) (x + y)n + (x – y)n = 2[nC0xn + nC2x

n – 2 + .......]= 2[Sum of terms at odd places]

last term nCn yn if n is even and nCn – 1xy n – 1 if n is odd

(e) (x + y)n – (x – y)n = 2[Sum of terms at even places]last term nCn – 1xyn – 1 if n is even and nCn y

n if n is odd

Illustration 1 :

In the expansion of (x + y)n, if the sum of odd terms be A and that of even terms be B, then 4AB equals.

(1) (x + y)n + (x – y)n (2) (x + y)n – (x – y)n

(3) (x + y)2n – (x – y)2n (4) (x + y)2n + (x – y)2n

Solution :

(x + y)n = A + B

(x – y)n = A – B

4AB = (A + B)2 – (A – B)2

= (x + y)2n – (x – y)2n

Hence Ans is (3)

General term in the expansion of (x + y)n

The (r + 1)th term = Tr + 1 is called the general term.

rrnrr yxnCT

1

Illustration 2 :

Find (i) coefficient of x9 and(ii) term independent of x

in the expansion of 9

2

31

x

x

Solution :

rrnr

nr CT term) second(term) first(1 in the expansion of

92

31

x

x

r

r

r

r

rrr

rr xCx

xCT 31899291 3

)1(31)1()(

(i) We want coefficient of x9

18 – 3r = 93r = 9 ; r = 3

Coefficient of x9 is 9

2831)1( 3

33

9 C

(ii) For the term independent of x, we want the power of x to be 0. 18 – 3r = 0 ; r = 6

24328

36789

3)1(

66

6

69

16

CT .


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AIEEE/State CETs Binomial Theorem and its Application

Middle term in the expansion of (x + y)n

Case 1.n = 2m ; m NThen, number of terms in the expansion = n + 1 = 2m + 1; Thus middle term is Tm + 1

i.e. th

12

n term

22

21 ,

nn

nn

m yxCT

Case 2.n = 2m + 1 ; m N

Then, number of terms in the expansion is = n + 1 = 2m + 2. There are two middle terms, viz. 21 and mm TT .

i.e. thn

21

and thn

23

term

21

21

211

nn

nn

m yxCT

21

21

212

nn

nn

m yxCT

Illustration 3 :

Find the middle term in the expansion of(i) (3x + 2y)10

(ii)932

x

x

Solution :

(i) Since 10 is even number hence middle term will be th

1

210

i.e. 6th term

T6 = 10C5(3x)10 – 5 (2y)5

= 5555 231.2.3.4.56.7.8.9.10 yx

= 7 × 27 × 37 × x5y5

(ii) Since 9 is odd number, so there will be two middle term i.e. th

219

and th

239

terms

i.e. 5th and 6th terms

T5 = 9C4(2x)5

43

x = 9C4.25.34.x

T6 = 9C5(2x)4

53

x = 9C5.24.35.x–1


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pth term from the end in the expansion of (x + y)n

pth term from end = (n + 2 – p)th term from begining

= )1(11

pnppn

n yxC

Illustration 4 :

Find the 4th term from end in the expansion of 9

2

3 22

xx

.

Solution :

4th term from end = (9 – 4 + 2) = 7th term from begining

i.e. 3

6

2

33

69

76722

2 xxxCT

Numerically greatest term in (1 + x)n

The method for finding the greatest term is as below :

1. Find the value of ||1

||)1(x

xnK

.

2. If K is an integer, then TK and TK + 1 both are equal and greatest.

3. If K is not an integer then T[K] + 1 is the greatest term where [K] is the greatest integral part of K.

4. To find the greatest term in (x + y)n = xn n

xy

1 , find the greatest term in

n

xy

1 and multiply with xn.

Illustration 5 :

Find the greatest term in the expansion of (2 + 3x)9 if 23

x .

Solution :In the expansion of (x + a)n

11

r

r

TT

ax

nr

1

1

13126

13909

4910

233

21

19

r

rmax. = 6

1rT6

696

9

2332

C 6

12

238

!3!6!9

123221

.


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Properties of Binomial CoefficientsIn the expansion of (1 + x)n, the coefficients, nC0, nC1, nC2 ..... nCn are denoted by C0, C1, C2 ....... Cn then.

1. If n is even, then greatest coefficient is nCn/2.

2. If n is odd, then greatest coefficients are nC(n –1)/2 or nC(n +1)/2

3. r.nCr = n.n – 1Cr – 1 0 < r n.

4. 111

1

nC

rC r

nr

n

5. )1(1

r

nr

n CrnC

6. rrn

nn

r

r

C

C 1

1

7. rrrC

nCC nn 1

1

8. yxnnyx CC or x + y = n

9. nCr = nCn – r

Note : (1) The number of terms in the expansion of (x + y + z)n is 2

)2()1( nn where n N.

(2) Sum of all coefficients in the expansion of (x1 + x2 + ..... + xK)n is obtained by keeping eachxi = 1 so it is kn.

Some Relation involving Binomial Coefficients(1 + x)n = 1 + nC1x + nC2x

2 + ......... + nCnxn …(1)

Relation (1) is important. Many properties of binomial coefficients are obtained by

(i) Giving different values to x in (1) and/or

(ii) Differentiating (1) with respect to x and then putting convenient values of x, and/or

(iii) Integrating (1) with respect to x and then putting convenient values of x.

Some important relation are given below :

1.n

n

n

rr

n CCCCC 2.......2100

2. C0 + C2 + C4 ......... = C1 + C3 + ........ = 2n – 1

3.1

3210or1

2.........32.

nn

nn

rr

n nCCCCCr

4. 112

1

1

0

nr

n nn

r

Cr

5. C0 – C1 + C2 – C3 + ......... + (–1)n Cn = 0


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6. 22

0

2221

20 )!(

!2)(.........n

nCCCCC nn

n

rr

nn

7.

evenisif.1)–(

oddisif,0..........)1(

2//2

223

22

21

20 nC

nCCCCC

nnn

nn

8. C0C1 + C1C2 + .......... + Cn – 1Cn = 2nCn – 1

Illustration 6 :

Show that C02 + 2C1

2 +.......+(n + 1)Cn2 = !)1–(!

)1–2()2(nn

nn

Solution :

Let S = C02 + 2C1

2 +......+(n + 1)Cn2

Also S = (n + 1)Cn2 + nCn

2–1 + (n – 1)C2

n–2 +.......+ C02

S + S = (n + 2)C02 + (n + 2) C1

2 +......+(n + 2)Cn2

2S = (n + 2)[C02 + C1

2 +......+Cn2]

2)!(!)2()2(

nnn

S !)1(!

!)12)(2(

nnnn

SOLVED EXAMPLES

Example 1 :

If the coefficients of rth, (r + 1)th and (r + 2)th terms in the expansion of (1 + x)14 are in A.P., find r.

Solution :

In the expansion of (1 + x)14, the rth, (r + 1)th and (r + 2)th terms are

11

14 r

rr xCT ; rrr xCT 14

1 ; 11

142

r

rr xCT

and the coefficients are 11414

114 and, rrr CCC respectively

)(2 14rC 1

141

14 rr CC

!)14(!!142

rr

!)13(!)1(!14

!)15(!)1(!14

rrrr

rrrrrr

)1(1

)14()15(1

)14(2

Multiplying throughout by r(r + 1) (14 – r) (15 – r), we have

222 29210)1514(2 rrrrrr

045142 rr ; r = 5 or 9


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Example 2 :

Evaluate (0.99)15 correct to four decimal places.

Solution :

(0.99)15 = (1 – 0.01)15

33

1522

151

15 )01.0()01.0()01.0(1 CCC

32 )01.0(455)01.0(105)01.0(151

000455.00105.015.01

1504.00105.1)99.0( 15

= 0.8601

Example 3 :

Find the coefficient of x4 in the expansion of 1132 )1( xxx

Solution :

112111132 )1()1()1( xxxxx

Terms in the expansion of (1 + x2)11 would involve even powers of x only.

Term involving x4 in (1 + x)11·(1 + x2)11 can arise in following ways :

x4 from (1 + x)11 and x0 from (1 + x2)11



Coefficient of x4 in 1132 )1( xxx is

211

011

111

211

011

411 CCCCCC

2101111

21011

24891011

= 330 + 605 + 55 = 990.

Example 4 :

Show that 6n + 2 + 72n + 1 is divisible by 43 Nn .

Solution :

122 76 nn

nn )49(7636 nn )643(76)743(

)643(767643 nnn k (write the binomial expansion of (43 + 6)n)

multiple of 43


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Example 5 :Prove the following :

(i)!)(!)(

!)2(.........22110 rnrnnCCCCCCCC nrnrrr

In particular, 2

221

20 )!(

!)2(.........n

nCCC n

(ii)!!2

!)2(2 1

0 nnnCC n

jinji

.

Solution :

(i) nx 2)1(

n

j

jj

n xC2

0

2 …(1)

nx 2)1( nn xx )1()1(

).....().....1( 110

221 n

nnnnnnn

nnn CxCxCxCxCxC …(2)

Coefficient of xn + r in (1) is 2nCn + r

Coefficient of xn + r in (2) is

nrnrrr CCCCCCCC .......22110

!)(!)(

!)2(........110 rnrnnCCCCCC nrnrr

.

(ii) nnCCC 2.......10 where r

nr CC

nnCCC 22

10 2).......(

nji

nji

n

ii CCC 2

01

2 22

21

)!(!22 2

2

0

nnCC n

njiji

aakash - binomial theorem and its application

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