binomial theorem - wordpress.com...ein binomial theorem i.e., r = 6. thus, the 7th term has x– 8...

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Binomial Theorem

1. DEFINITION OF BINOMIAL EXPRESSIONAND BINOMIAL EXPANSION :An expression containing two terms, is called a binomial expression. For examplea + b/x, x + 1/y, a – y2 etc. are binomial expressions. Expansion of (x + a)n is called BinomialExpansion.Expression containing three terms are called trinomials. For example x + y + z is a trinomial expression.In general an expression containing more than two terms is called a multinomial.

1.1 Definition of binomial theorem :If n is a positive integer and x, y are two complex numbers, then

n

n n n r rr

r 0

x y C x y

= nC0xn + nC

1xn – 1 y + nC

2xn – 2 y2 + . . . + nC

n yn . . . (i)

The coefficients nC0, nC

1, . . . , nC

n are called binomial coefficients, while (i) is called the binomial

expansion.

1.2 Some Important Facts Regarding Binomial Expansion :

(i) There are (n + 1) terms in the expansion.

(ii) The sum of the exponents of x and y in any term of the expansion is equal to n.

(iii) The binomial coefficients of terms equidistant from the beginning and the end are equal,since nC

r = nCn – r .

(iv) The term nCr xn – r yr is the (r + 1)th term from the beginning of the expansion. It is usually

denoted by Tr + 1

and is called the general term of the expansion.

(v) The rth term from the end is equal to the (n – r + 2)th term from the beginning, i.e.,nCn – r + 1

xr – 1 yn – r + 1 .

(vi) If n is even, then the expansion has only one middle term, then

12

th term i.e.,

n n / 2 n / 2n / 2C x y .

If n is odd, then the expansion has two middle terms, then 1

2

th term and then 3

2

th

term i.e., n 1 / 2 n 1 / 2n

n 1 / 2C x y and

n 1 / 2 n 1 / 2nn 1 / 2C x y .

Illustration 1 :

Find the coefficient of x–8 in the expansion of10

2

1x

2x

.

Solution:We have

Tr + 1

= 10Cr(x)10 – r

r

2

1

2x

= 10Cr (– 1)r 2– r x10 – 3r

To find the coefficient of x–8, we have10 – 3r = – 8

Binomial Theorem

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i.e., r = 6.Thus, the 7th term has x– 8 and its coefficient is

610 66

105C 1 2

32 .

Illustration 2:

Find the 3rd term from the end in the expansion of9

3 1x

x

.

Solution:The 3rd term from the end is equal to (9 – 3 + 2)th term, i.e., the 8th term from the beginning. Hence,the required term is

T8 = 9C

7 (x3)9 – 7

71 9.8 1 36

x 2 x x

.

Illustration 3:

Find the middle term in the expansion of9

bax

x

.

Solution:The expansion has two middle terms,

viz9 1

2

th = 5th term and9 3

2

th = 6th term.

Hence, the middle terms are

T5 = 9C

4 (ax)9 – 4

45 4 5 4b 9.8.7.6

a b x 126a b xx 1.2.3.4

and T6 = 9C

5 (ax)9 – 5

54 5 1 4 5 1b 9.8.7.6

a b x 126a b xx 1.2.3.4

drill exercise - 1

1. If p and q be positive, then prove that the coefficients of xp and xq in the expansion of (1 + x)p+ q willbe equal.

2. (a) Find the term independent of y in the expansion of (y–1/6 - y1/3)9 .

(b) Find the coefficient of x5 in the expansion of the product (1 + 2x)6 (1 – x)7 .

3. Find the number of integral terms in the expansion of 8732 3 .

4. If coefficient of (2r + 3)th and (r - 1)th terms in the expansion of (1 + x)15 are equal, then find thevalue of r.

5. (a) Find the middle term in the expansion of12

bxx

a

.


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(b) Find the middle term in the expansion of

93

6

xx3

.

2. SOME STANDARD EXPANSIONS :

(i) Consider the expansion n

n n n r rr

r 0

x y C x y

. . . (i)

If we replace y by – y in equation (i), we have

n

n rn n r rr

r 0

x y C 1 x y

. . . (ii)

= nn n n n 1 n n 2 2 n n0 1 2 nC x C x y C x y C 1 y . . . (ii)

(ii) Adding equations (i) and (ii), we have

n nn n n n 2 2 n n 4 40 2 4

1C x C x y C x y x y x y

2 . . . (iii)

and substracting equations (ii) from (i) we have,

n nn n 1 n n 3 3 n n 5 51 3 5

1C x y C x y C x y x y x y

2 . . . (iv)

(iii) Putting x = 1 and y = 1 in equation (i), we have

n n n n n n0 1 2 n 1 nC C C C C 2 . . . (v)

Thus, we see that the sum of the binomial coefficients of (x + y)n is 2n.

(iv) Putting x = 1 and y = 1 in equation (iii) and (iv), we have

n n n n 1 n n n0 2 4 1 3 5C C C 2 C C C . . . (vi)

(v) Putting x = 1 and replacing y by x in equation (i), we have(1 + x)n = nC

0 + nC

1x + nC

2x2 + . . .+ nC

nxn . . . (vii)

Replacing x by – x in equation (vii), we have(1 – x)n = nC

0 – nC

1 x + nC

2x2 – . . . + nC

n (– 1)n xn . . . (viii)

2.1 Important Formulae :

If C C C C0 1 2 3, , ,..........., represent n n n nC C C C0 1 2 3, , ..........., in the expansion of (1 + x)n . Then

(i) C C C C Cnn

0 1 2 3 2 ...........

(ii) C C C C Cnn0 1 2 3 1 0 ............. ( )


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(iii) C C C C C C n0 2 4 1 3 5

12 ............. ...........

drill exercise - 2

1. (a) Find the value of 2 1 2 16 6

e j e j . (b) Find the sum of the series

10

0

20

rrC .

2. Find the value of 1114

514

314

114 CCCC .

3. (a) Find the sum of the series

1

0 1

n

r rn

rn

rn

CC

C. (b) Find the value of

nr

mr m

C ,n m

.

4. Find the sum of the series

n

0rr4

r

r3

r

r2

r

rrnr to...

2

15

2

7

2

3

2

1C)1(

5. If the 4th term in the expansion ofn

x

1ax

is 5/2, then find the values of a and n.

2.2 Questions related with integral and fractional part

Illustration 4 :If (7 + 4 3 )n = I + F, where I is a positive integer and F is a proper fraction, then show that

(I + F) (1 – F) = 1.Solution :

Let G = (7 – 4 3 )n.Clearly, if we add G and I + F, we get an integer i.e.,

I + F + G = (7 + 4 3 )n + (7 – 4 3 )n

= 2[nC07n + nC

27n–2 (4 3 )2 + . . . ] = an even integer..

F + G = 1 G = 1 – FHence, (I + F) (1 – F) = (I + F) G = (7 + 4 3 )n (7 – 4 3 )n = 1.

drill exercise 3

1. Find the greatest integer less than or equal to ( 2 + 1)6.

2. If n is a natural number, show that the integral part of (5 + 2 6 )n is odd natural number..

3. Show that the integer just greater than ( 3 + 1)2m contains 2m+1 as a factor..

4. If (6 6 + 14)2n + 1 = R and F = R – [R], where [R] denotes the greatest integer less than or equal toR, prove that RG = 20 2n+1.


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5. If (5 + 6 )n = I + f, where I and n are positive integers and f is a positive fraction less than one, show

that (I + f) (1 – f) = 1.

3. GREATEST BINOMIAL COEFFICIENT :The greatest coefficient depends upon the value of n. n no. of greatest coefficient (s) Greatest coefficientEven 1 nC

n/2

Odd 2 n 1

n

2

C and n 1

n

2

C

(Values of both these coefficients are equal)Clearly greatest binomial coefficient corresponds to the coefficient of middle term.

4. NUMERICALLY GREATEST TERM OF BINOMIAL EXPANSION :(a + x)n = C

0an + C

1an – 1 x + . . . C

n – 1 a xn – 1 + Cnxn

nr 1 r

nr r 1

T C x n r 1 x

T a r aC

Ifn r 1 x

1r a

, for given a, x and n, then r n 1

a1

x

So numerically greatest term will be Tr + 1

, where r =n 1

a1

x

[ ] denotes the greatest integer function.

Note : Ifn 1

a1

x

itself is a natural number, then TT

r = T

r + 1 and both the terms are numerically greatest term.

Illustration 5 .

Given that the 4th term in the expansion of10

3x2

8

has the maximum numerical value, find the

range of values of x for which this will be true.

Solution:According to the question, |t

4| |t

3|, |t

4| |t

5|.

Now tr + 1

= 10Cr. 210 – r

r3x

8

t4 = 10C

3. 27.

33x

8

, t3 = 10C

2. 28.

23x

8

and t5 = 10C

4. 26.

43x

8


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Now, |t4| |t

3|

10C3. 27.

33

8

. | x |3 10C2. 28.

23

8

. | x |2 . . . (i)

and |t4| |t

5|

10C3. 27.

33

8

. |x|3 10C4. 26.

43

8

. | x |4 . . . (ii)

from (i)

3 210.9.8 3 10.9. . | x | .2 | x |

6 8 2

or 45| x |3 90 | x |2

or | x |3 – 2|x|2 0

or | x | 2 ,as x can not be zero.

from (ii)3 410.9.8 10.9.8.7 3

.2 | x | . . | x |6 24 8

or | x |3 47 3. | x |

8 8

or | x |321

1 | x | 064

211 | x | 0

64

| x | 64

21

Thus, we get | x | 2 and | x | 64 1

321 21 . So x

64 64, 2 2,

21 21

drill exercise 4

1. Prove that the greatest coefficient in the expansion of (1 + x)2n is double the greatest coefficient in theexpansion (1 + x)2n – 1.

2. Find numerically the greatest term in the expansion of (3 – 5x)15 when x = 1/5.

3. Find the greatest term in the expansion of (1 + x)10 when x = 2/3.

4. If the expansion ofn

3

x

2

3

when x =

2

1, it is known that 6th term is the greatest term, then find the

possible positive integral values of n.


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5. Show that if the greatest term in the expansion of (1 + x)2n has also the greatest coefficient, then x lies

between1n

n

and

n

1n .

5. SERIES OF BINOMIAL COEFFICIENT :

5 . 1 Sum of the series by the use of differentiation :Generally we use the method of differentiation when the coefficient of binomial expansion C

k is a

polynomial in k

Important Formulae :


(i) C C C nC nnn

1 2 312 3 2 ............. .

(ii) C C C n Cnn1 2 3

12 3 1 0 ............. ( ) .

Illustration 6:Find the sum of the series

C0 – 3C

1 + 5C

2 + . . . + (– 1)n (2n + 1)C

n.

Solution :We have

(1 – x)n = C0 – C

1x + C

2x2 – . . . + C

n (– 1)nxn . . . (i)

Replacing x by x2 in equation (i), we have(1 – x2)n = C

0 – C

1x2 + C

2x4 – . . . + C

n (– 1)n x2n . . . (ii)

Multiplying throughout by x, we havex(1 – x2)n = C

0 x – C

1 x3 + C

2x5 – . . . + C

n (– 1)n x2n + 1 . . . (iii)

Differentiating equation (iii) w.r.t. x, we have

(1 – x2)n – 2nx2 (1 – x2)n – 1 = C0 – 3C

1x2 + 5C

2x4 – . . . + n 2n

n1 2n 1 C x . . . (iv)

Putting x = 1 in equation (iv), we haveC

0 – 3C

1 + 5C

2 – . . . + (– 1)n (2n + 1) C

n = 0.


12. C1 + 22. C

2 + 32. C

3 + . . . + n2. C

n

Solution :We have

(1 + x)n = C0 + C

1x + C

2x2 + . . . + C

nxn . . . (i)

Differentiating equation (i) w.r.t. x, we haven(1 + x)n – 1 = C

1 + 2C

2x + 3C

3x2 + . . . + nC

nxn – 1 . . . (ii)

Multiplying equation (ii) throughout by x, we havenx(1 + x)n – 1 = C

1x + 2C

2x2 + 3C

3x3 + . . . + nC

nxn . . . (iii)

Differentiating equation (iii) w.r.t. x, we have


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n(1 + x)n – 1 + n (n – 1)(1 + x)n – 2

= C1 + 22. C

2x + 32. C

3x2 + . . . + n2. C

n xn – 1 . . . (iv)

Putting x = 1 in equation (iii), we have12. C

1 + 22. C

2 + 32. C

3 + . . . + n2. C

n

= n. 2n – 1 + n(n – 1). 2n – 2 = (n2 + n) 2n – 2

= n (n + 1)2n – 2.5.2 Sum of the series by the use of integration :

Generally we use integration for the series having terms of the form m kCr

m 1 or of the form

m kC

rm 1 m 2 . . . m j

.


aC0 + a2 3 n 11 2 nC C C

a . . . a2 3 n 1

Solution :We have (1 + x)n = C

0 + C

1x + C

2x2 + . . . + C

nxn . . . (i)

Integrating equation (i) w.r.t. x, we have

a a

n 2 n0 1 2 n

0 0

1 x dx C C x C x . . . C x dx . . . (ii)

Now, L.H.S. =

an 1 n 1

0

1 x 1 a 1

n 1 n 1

and R.H.S =

a2 32 31 2

0 1 2 0

0

C Cx xC C C . . . aC a a . . .

2 3 2 3

Hence, we have

aC0 + a2

n 131 2 1 a 1C C

a . . .2 3 n 1

Illustration 9:

Find the sum of the series C1 – n 132 4 nCC C C

. . . 12 3 4 n

.

Solution :We have (1 – x)n = C

0 – C

1x + C

2x2 – . . . + C

n (– 1)n xn

i.e. n n 12 3 n1 2 3 n1 1 x C x C x C x . . . C 1 x

i.e.

nn 12 n 1

1 2 3 n1 1 x

C C x C x . . . C 1 xx

. . . (i)

Integrating equation (i) w.r.t. x from 0 to 1, we have


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n1 12

1 2 30 0

1 1 xdx C C x C x . . . dx

x

Now, L.H.S. =

1 n

0

1 xdx

1 x

=

11 2 3 n2 n 1

0 0

x x x1 x x . . . x dx x . . .

2 3 n

= 1 +1 1 1

. . .2 3 n

and R.H.S.=

12 332

1 2 3 1

0

CCx xC x C C . . . C . . .

2 3 2 3

+ (– 1) n 1 nC

n 1

Hence, we have

C1 – n 132 nCC C 1 1 1

. . . 1 1 . . .2 3 n 2 3 n

.

drill exercise 51. If C

0, C

1, C

2, . . . C

n denote the coefficient in the binomial expansion of (1 + x)n, then prove that :

C0 – C

1 + C

2 – C

3 + . . . + (–1)n C

n = 0

2. If C0, C

1, C

2, . . . C


C0 + 3C

1 + 5C

2 + . . . (2n + 1) C

n = (n + 1).2n

3. If C0, C

1, C

2, . . . C


(1.2) C2 + (2.3) C

3 + . . . + ((n – 1).n) C

n = n (n – 1)2n–2

4. If Cr = nC

r then prove that

2

C1 +4

C3 +6

C5 + . . . =1n

12n

.

5. If C0, C

1, C

2, . . . C


3C0 +

2

32

C1 +

3

33

C2 +

4

34

C3 + . . . +

1n

3 1n

Cn =

1n

14 1n

.

6. If (1 + x)n =

n

0r

nr

n xC , then prove that

2.1

C0 .22 +3.2

C1 .23 +4.3

C2 .24 + . . . + )2n)(1n(

Cn

2n+2 =)2n)(1n(

5n23 2n

5.3 Binomial Theorem For Any Rational Index :

(1 + x)n = 11

2

1 2

32 3

nx

n nx

n n nx

( )

!

( )( )

! where n R , -1 < x < 1.

Deduction :

(1 - x)-1 = 1 + x + x2 + x3 +.....................+ xr ................ x 1

(1 - x)-2 = 1 + 2x + 3x2 +.........................+ r 1 x ............ x 1r b g


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(1 - x)-3 = 1 + 3x + 6x2 + 10 x3 +.................+(r +1)(r + 2)x

2

r

.............. x 1

5.4 Multinomial Expansion ( )n N :

General terms in expansion of x x x xk

n

1 2 3 .........d i is

n!

a ! a ! a !.........a !.x .x .x ........x

1 2 3 k1

a2

a3

ak

ak1 2 3

where

a a a a n a n i kk i1 2 3 0 1 2 3 .......... , , , , ,....... and the number of terms in the expansion aren k

kC

11.

Number of terms in (x + y)n = n C1 1

Number of terms in ( )x y z Cn n 22

Number of terms in ( )x y z Cn 33

5.5 Sum of the series by comparing the coefficients of some power of x in an expansion :In this method we use the fact that coefficient of same power of x in an appropriate identity is the givenseries.

Important Formulae :


(i) C C C C Cnn

n02

12

22 2 2 .............

(ii) C C C C C C C C C or Cr r r n r nn

n rn

n r0 1 1 2 22 2 .............


mCr + mCr – 1

nC1+ mCr – 2

nC2 + . . . + nC

r

where r < m, n and m, n, r N.

Solution :We have

(1 + x)n = nC0 + nC

1 x + nC

2x2 + . . . + nC

rxr + . . . + nC

n xn . . . (ii)

and (1 + x)m = mC0 + mC

1x + . . . + mCr – 2 xr – 2 + mCr – 1 x

r – 1

+ mCr xr + . . . + mC

mxm . . . (ii)

Hence, the given series= coefficient of xr in (1 + x)n (1 + x)m

= coefficient of xr in (1 + x)m + n = m nrC .

Illustration 11:

Find the sum of the series 2 2 2 21 2 3 nC 2.C 3.C . . . n.C

Solution :We have (1 + x)n = C

0 + C

1x + C

2x2 + . . . + n. C

nxn . . . (i)

andn

0 1 2 n2 n

1 1 1 11 C C C . . . C

x x x x

. . . (ii)

Differentiating equation (i) w.r.t. x, we have


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n(1 + x)n –1 = C1 + 2C

2x + 3C

3x2 + . . . + nC

n xn – 1 . . . (iii)

Hence, the given series = coefficient of x– 1 inn

11

x

. n (1 + x)n – 1

= n × coefficient of xn – 1 in (1 + x)2n – 1

= 2n 1n 1n. C .

5.6 Sum of the series by equating the real and imaginary parts :Illustration 12:

Prove that C1 – C

3 + C

5 – . . . = 2n/2 sin

n

4

Solution :Consider the expansion(1 + x)n = C

0 + C

1x + C

2x2 + . . . + C

nxn . . . (i)

Putting x = – i in equation (i) we have(1 – i)n = C

0 – C

1i – C

2+ C

3i + C

4 + . . . (– 1)n C

nin

or (2)n/2n n

cos i sin4 4

= (C

0 – C

2 + C

4 – . . . ) – i. (C

1 – C

3 + C

5 – . . . ) . . . (ii)

equating the imaginary part in (i), we get

C1 – C

3 + C

5 – . . . = 2n/2

nsin

4

Illustration 13:If (3 + 4x)n = p

0 + p

1x + p

2x2 + p

3x3 + . . . + p

nxn, then prove that

(p0 – p

2 + p

4 – . . . )2 + (p

1 – p

3 + p

5 – . . .)2 = 25n

Solution :Consider the expansion (3 + 4x)n = p

0 + p

1x + p

2x2 + p

3x3 + . . . + p

nxn

Putting x = i in the above expansion we get,(3 + 4i)n = (p

0 – p

2+ p

4– . . .) + i (p

1 – p

3 + p

5 – . . .)

Equating the square of the modulus, we get,(p

0 – p

2 + p

4 – . . . )2 + (p

1 – p

3 + p

5 – . . .)2 = 25n

drill exercise 61. If C

0, C

1, C

2, C

3, . . ., Cn–1, Cn

denote the binomial coefficients in the expansion of (1 + x)n, thenprove that :

C0C

1 + C

1C

2 + C

2C

3 + . . . + Cn–1 .Cn

=)!1n()!1n(

)!n2(

=

)!1n(

)1n2(...5.3.1

.n.2n

2. If C0, C

1, C

2, C

3, . . ., Cn–1, Cn

denote the binomial coefficients in the expansion of (1 + x)n, thenprove that :

C0C

r + C

1C

r+1 + C

2C

r+2 + . . . + Cn–r C

n = )!rn()!rn(

)!n2(

3. If C0, C

1, C

2, C

3, . . ., Cn–1, Cn

denote the binomial coefficients in the expansion of (1 + x)n, then find

the value of C12 +

2

21 C

22 +

3

321 C

32 + . . . +

n

n...321 C

n2.


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4. Prove that :n–1Cn–1.

nC1 + n–1Cn–2.

nC2 + n–1Cn–3.

nC3 + . . . + n–1C

0.nC

n = 2n–1Cn–1

5. Prove that :n–1C

0.nC

2 + n–1C

1.nC

3 + n–1C

2.nC

4 + . . . + n–1Cn–2.

nCn = 2n–1Cn–2

5.7 Approximations :

(1 + x)n = .....x3.2.1

)2n)(1n(nx

2.1

)1n(nnx1 32

If x < 1, the terms of the above expansion go on decreasing and if x be very small, a stage may bereached when we may neglect the terms containing higher powers of x in the expansion. Thus, ifx be so small that its squares and higher powers may be neglected then (1 + x)n = 1 + nx,approximately . This is an approximate value of (1 + x)n .

5.8 Exponential Series :

(i) ex = .........!3

x

!2

x

!1

x1

32

;where x may be any real or complex and e =n

n n

11Lim

(ii) ax = .......an!3

xan

!2

xna

!1

x1 3

32

2

where a > 0

Note : (a) e = .......!3

1

!2

1

!1

11

(b) e is an irrational number lying between 2.7 and 2.8. Its value correct upto 10 place ofdecimal is 2.7182818284.

(c) e + e-1 =

........

!6

1

!4

1

!2

112

(d) e - e-1 =

........

!7

1

!5

1

!3

112

(e) Logarithms to the base ‘e’ are known as the Napierian system, so named after Napier,their inventor. They are also called Natural Logarithm.

5.9 Logarithmic Series :

(i) ........4

x

3

x

2

xx)x1(n

432

where 1x1

(ii) ..........4

x

3

x

2

xx)x1(n

432

where 1x1

(iii) 1|x|.......5

x

3

xx2

)x1(

)x1(n

53

Remember :


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Study

Steps

.in

Binomial Theorem

(a) 2n..........4

1

3

1

2

11

(b) xe xn

(c) 693.02n (d) 303.210n

drill exercise - 7

1. Find the coefficient of xn in the series .......!3

)bxa(

!2

)bxa(

!1

bxa1

32

2. If x is so small that is square and higher powers may be neglected, then prove that :

x24

411

)x4(

)x1()x31(2/1

3/52/1

.

3. Ifx1

ex

= AA

0 + A

1x + A

2x2 + . . . + A

nxn + . . . , then prove that

(a) A0 =1 (b) A

n – An–1 = !n

1

4. Prove that :3

1 + 33.3

1 + 53.5

1 + 73.7

1 + . . . =

2

1log 2

5. Prove that : 1 +

3

1

2

1.4

1 +

5

1

4

1. 24

1 +

7

1

6

1. 34

1 + . . . = log 12


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Study

Steps

.in

Binomial Theorem

Answer Key

drill exercise - 1

2. (a) –84 (b) 171 3. 15 4. 5

5. (a) 924a6b6 (b) T5 =

17x8

189, T

6 = – 19x

16

21

drill exercise - 2

1. (a) 140 2 (b) 102019

2

12 C 2. 213 - 14

3. (a)2

n(b) n + 1C

m + 14.

12

1n

5. 1/2, 6

drill exercise - 3

1. 197

drill exercise - 4

2. T4 = 455 × 312 and T

5 = 455 × 312 3.

4

3

2210

4. n = 49, 50, 51, . . . 59

drill exercise - 6

3.2

1

1C

2

n1 n

n2

drill exercise - 7

1.!n

b.e na


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