Binomial Theorem

MATHS

Binomial Theorem Binomial expression : Any algebraic expression which contains two dissimilar terms is called binomial expression. 1

For example : x + y, x 2y +

xy

2

, 3 – x,

x2  1 +

1 ( x  1)1/ 3 3

etc.

Terminology used in binomial theorem : Factorial notation :

or n! is pronounced as factorial n and is defined as

n(n  1)(n  2)........ 3 . 2 .1 ; if n  N n! =  1 ; if n  0 

n

Note : n! = n . (n – 1)! ;

N

Mathematical meaning of nC r : The term nCr denotes number of combinations of r things choosen from n distinct things mathematically, nCr =

n! , n N, r  W, 0 r n (n  r )! r!

n Note : Other symbols of of nCr are   and C(n, r). r 

Properties related to n C r : (i)

n

Cr = nCn – r

Note : If nCx = nCy (ii)

n

Either x = y or

C r + nC r – 1 = n

(iii)



n

Cr

Cr 1

=

n r

n+1

x+y=n

Cr

nr 1 r

(v)

If n and r are relatively prime, then nCr is divisible by n. But converse is not necessarily true.

Cr–1 =

n–2

n(n  1)(n  2).........(n  (r  1)) r (r  1)(r  2).......2 .1

n

Cr =

n–1

n(n  1) r(r  1)

(iv)

Cr–2 = ............. =

Statement of binomial theorem : (a + b)n = nC0 anb0 + nC1 an–1 b1 + nC2 an–2 b2 +...+ nCr an–r br +...... + nCn a0 bn

"manishkumarphysics.in"

1

MATHS where n  N n

or

n

(a + b) =



n

C r a n r b r

r0

Note : If we put a = 1 and b = x in the above binomial expansion, then or (1 + x)n = nC0 + nC1 x + nC2 x 2 +... + nCr x r +...+ nCn x n n

or

n

(1 + x) =



n

Cr x r

r 0

Example # 1 : Expand the following binomials :

Solution :

 3x 2  1     2  

4

(i)

(x – 3)5

(i)

(x – 3)5 = 5C0x 5 + 5C1x 4 (– 3)1 + 5C2 x 3 (– 3)2 + 5C3 x 2 (–3)3 + 5C4 x (– 3)4 + 5C5 (– 3)5 5 4 = x – 15x + 90x 3 – 270x 2 + 405x – 243

(ii)

4

(ii)

 3x 2  1   = 4C + 4C  0 1 2  

 3x 2    4  2  + C2   3

 3x 2   + 4C + 4C3    4 2   = 1 – 6x 2 +

Solution :

20

 2x   = C0   3  20

20

up to four terms

 2x  + C1    3 

19

20

 2x  + C3    3  20

18

2 + 20.   3

4

20

20

 2x   =   3 

 3x 2      2  

2

81 8 27 4 27 6 x – x + x 16 2 2

 2x 3y    Example # 2 : Expand the binomial  2   3  2x 3y     2   3

 3x 2     2  

16

2 x 19y + 190 .   3

17

 3y   2x    + 20C   2 2    3 

 3y     2 

18

 3y     2 

2

3

+ .... 14

2 x 18 y2 + 1140   3

x 17 y3 + .....

Self practice problems 6

(1)

y  Write the first three terms in the expansion of  2   . 3  


2

MATHS 5

(2)

 x2 3    . Expand the binomial  3 x  

Answers :

(1)

64 – 64y +

80 2 y 3

(2)

135 243 x10 5 7 10 4 + x + x + 30x + 2 + . 3 243 27 x x5

Observations : (i) (ii) (iii) (iv)

The number of terms in the binomial expansion (a + b) n is n + 1. The sum of the indices of a and b in each term is n. The binomial coefficients ( nC0, nC1 ..........nCn) of the terms equidistant from the beginning and the end are equal, i.e. nC0 = nCn, nC1 = nCn–1 etc. { nCr = nCn–r} The binomial coefficient can be remembered with the help of the following pascal’s Triangle (also known as Meru Prastra provided by Pingla)

Regarding Pascal’s Triangle, we note the following : (a) Each row of the triangle begins with 1 and ends with 1. (b) Any entry in a row is the sum of two entries in the preceding row, one on the immediate left and the other on the immediate right. Example # 3 : The number of dissimilar terms in the expansion of (1 – 3x + 3x 2 – x 3)20 is (A) 21 (B) 31 (C) 41 (D) 61 Solution :

(1 – 3x + 3x 2 – x 3)20 = [(1 – x)3]20 = (1 – x)60 Therefore number of dissimilar terms in the expansion of (1 – 3x + 3x 2 – x 3)20 is 61.

General term : (x + y)n = nC0 x n y0 + nC1 x n–1 y1 + ...........+ nCr x n–r yr + ..........+ nCn x 0 yn (r + 1)th term is called general term and denoted by T r+1. T r+1 = nCr x n–r yr Note : The rth term from the end is equal to the (n – r + 2) th term from the begining, i.e.

28th term of (5x + 8y)30

Example # 4 : Find

(i)

Solution :

T 27 + 1 = 30C27 (5x)30– 27 (8y)27 =

(i)

(ii)

n

Cn – r + 1 x r – 1 yn – r + 1

 4x 5    7th term of   5 2x 

9

30 ! (5x)3 . (8y)27 3 ! 27 !


3

MATHS (ii)

 4x 5    7th term of   5 2x   4x   T 6 + 1 = 9C 6   5 

96

9

 5     2x 

6

9!  4x   = 3!6!   5 

3

 5     2x 

6

10500

=

x3

Example # 5 : Find the number of rational terms in the expansion of (91/4 + 81/6)1000. Solution :



The general term in the expansion of 91/ 4  81/ 6 1000 r

Tr+1

 1  4 1000 = Cr  9   

r

 1 8 6    =  

1000

Cr 3



1000

1000 r 2

is

r

22

The above term will be rational if exponent of 3 and 2 are integers It means

1000  r r and must be integers 2 2

The possible set of values of r is {0, 2, 4, ............, 1000} Hence, number of rational terms is 501

Middle term(s) : (a)

n 2  If n is even, there is only one middle term, which is   2 

(b)

 n  1  If n is odd, there are two middle terms, which are   2 

th

th

term.

n 1   1 and   2 

th

terms.

Example # 6 : Find the middle term(s) in the expansion of 14

(i)

2   1  x   2  

(i)

2   1  x   2  

(ii)

3   3a  a  6 

   

9

14

Solution :

 14  2  th  term. Here, n is even, therefore middle term is   2 

It means T 8 is middle term 7

 x2  429 14  =– T 8 = 14C7   x .  16  2 

(ii)

3   3a  a  6 

   

9


4

MATHS  9  1 9 1   1 th.  th &  Here, n is odd therefore, middle terms are   2   2 

It means T 5 & T 6 are middle terms 4

9

9–4

 a3   6 

  = 189 a17  8 

9

9–5

 a3   6 

  = – 21 a19.  16 

T 5 = C4 (3a)

T 6 = C5 (3a)

5

 4 1  Example # 7 : Find the coefficient of x 32 and x –17 in  x  3  x   Solution :

15

.

Let (r + 1)th term contains x m r

Tr + 1 (i)

(ii)

 1  = 15Cr (x 4)15 – r   3  = 15Cr x 60 – 7r (– 1)r  x 

for x 32 , 60 – 7r = 32  7r = 28  r = 4, so 5th term. 15 32 4 T 5 = C4 x (– 1) Hence, coefficient of x 32 is 1365 for x –17, 60 – 7r = – 17  r = 11 , so 12th term. T12= 15C11 x–17 (– 1)11 Hence, coefficient of x –17 is – 1365

Numerically greatest term in the expansion of (a + b) n , n  N Binomial expansion of (a + b) n is as follows : – (a + b)n = nC0 anb0 + nC1 an–1 b1 + nC2 an–2 b2 +...+ nCr an–r br +...... + nCn a0 bn If we put certain values of a and b in RHS, then each term of binomial expansion will have certain value. The term having numerically greatest value is said to be numerically greatest term. Let T r and T r+1 be the rth and (r + 1)th terms respectively Tr = nCr–1 an–(r–1) br–1 Tr+1 = nCr an–r br Now,

Consider

Tr 1 Tr

n

=

n

a n r b r

Cr

Cr 1 a

Tr 1 Tr

n r 1 r 1

b

=

nr 1 . r

b a

1

 n  r  1   r  

n 1 –1 r

b a

1 a b


5

MATHS n 1 r  1 a b

Case - 

(i) (ii) (iii)

When

n 1 a 1 b

is an integer (say m), then

T r+1 > T r when r < m (r = 1, 2, 3 ...., m – 1) i.e. T 2 > T 1, T 3 > T 2, ......., T m > T m–1 T r+1 = T r when r = m i.e. T m+1 = Tm T r+1 < T r when r > m (r = m + 1, m + 2, ..........n ) i.e. T m+2 < T m+1 , T m+3 < T m+2 , ..........T n+1 < T n

Conclusion : When

n 1 a 1 b

is an integer, say m, then T m and T m+1 will be numerically greatest terms (both terms are

equal in magnitude) Case -  When

(i)

n 1 a 1 b

is not an integer (Let its integral part be m), then

T r+1 > T r

i.e. (ii)

r<

n 1 a 1 b

(r = 1, 2, 3,........, m–1, m)

T 2 > T 1 , T 3 > T 2, .............., T m+1 > T m

T r+1 < T r

i.e.

when

when r >

n 1 a 1 b

(r = m + 1, m + 2, ..............n)

T m+2 < T m+1 , T m+3 < T m+2 , .............., T n +1 < T n

Conclusion : When

n 1 a 1 b

is not an integer and its integral part is m, then T m+1 will be the numerically greatest

term. Note : (i) If

In any binomial expansion, the middle term(s) has greatest binomial coefficient. In the expansion of (a + b) n n No. of greatest binomial coefficient Greatest binomial coefficient n Even 1 Cn/2 n Odd 2 C(n – 1)/2 and nC(n + 1)/2 (Values of both these coefficients are equal )


6

MATHS (ii)

In order to obtain the term having numerically greatest coefficient, put a = b = 1, and proceed as discussed above.

Example # 8 : Find the numerically greatest term in the expansion of (3 – 5x) 15 when x = Solution :

1 . 5

Let rth and (r + 1)th be two consecutive terms in the expansion of (3 – 5x) 15 Tr + 1  Tr 15 Cr 315 – r (| – 5x|)r  15Cr – 1 315 – (r – 1) (|– 5x|)r – 1

15 )! 3. 15 )! |– 5x |  (15  r ) ! r ! (16  r ) ! (r  1) ! 5.

1 (16 – r) 3r 5

16 – r  3r 4r  16 r4 Explanation: For r  4, T r + 1  T r 

T2 > T1 T3 > T2 T4 > T3 T5 = T4

For r > 5, T r + 1 < T r T6 < T5 T7 < T6 and so on Hence, T 4 and T 5 are numerically greatest terms and both are equal. Self practice problems : (3)

(4)

 2 3 Find the term independent of x in  x   x 

9

The sum of all rational terms in the expansion of (31/5 + 21/3)15 is (A) 60 (B) 59 (C) 95

(D) 105

8

(5)

1  Find the coefficient of x in (1 + 3x + x ) 1   x 

(6)

Find the middle term(s) in the expansion of (1 + 3x + 3x 2 + x 3)2n

(7)

Find the numerically greatest term in the expansion of (7 – 5x) 11 when x =

–1

Answers :

2

4

(3)

28.37

(4)

B

(5)

(6)

6n

(7)

T4 = –

440 × 7 8 × 5 3. 9

C3n . x3n

2 . 3

232


7

MATHS Result :

If ( A  B)n =  + f, where  and n are positive integers, n being odd and 0 < f < 1, then ( + f) f = k n where A – B2 = k > 0 and

A – B < 1. If n is an even integer, then ( + f) (1 – f) = k n

Example # 9 : If n is positive integer, then prove that the integral part of (7 + 4 3 )n is an odd number.. Solution :

Let

(7 + 4 3 )n =  + f

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

where  & f are its integral and fractional parts respectively. It means 0 < f < 1 Now,

0<7–4 3 <1 0 < (7 – 4 3 )n < 1

Let

(7 – 4 3 )n = f



0 < f < 1 Adding (i) and (ii)

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

 + f + f = (7 + 4 3 )n + (7 – 4 3 )n = 2 [nC0 7n + nC2 7n – 2 (4 3 )2 + ..........]  + f + f = even integer  (f + f must be an integer) 0 < f + f < 2  f + f = 1  + 1 = even integer therefore  is an odd integer. Example # 10 : Show that the integer just above ( 3 + 1)2n is divisible by 2n + 1 for all n  N. Solution :

Let ( 3 + 1)2n = (4 + 2 3 )n = 2n (2 +

3 )n =  + f

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

where  and f are its integral & fractional parts respectively 0 < f < 1. Now

0<

3 –1<1

0 < ( 3 – 1)2n < 1 Let

( 3 – 1)2n = (4 – 2 3 )n = 2n (2 –

3 )n = f.

........(ii)

0 < f < 1 adding (i) and (ii)  + f + f = ( 3 + 1)2n + ( 3 – 1)2n = 2n [(2 +

3 )n + (2 –

3 )n] = 2.2n [nC0 2n + nC2 2n – 2 ( 3 )2 + ........]

 + f + f =2n + 1 k (where k is a positive integer) 0 < f + f < 2  f + f= 1 n+1 +1=2 k.  + 1 is the integer just above ( 3 + 1)2n and which is divisible by 2n + 1. Example # 11 : Show that 9n + 7 is divisible by 8, where n is a positive integer. 9n + 7 = (1 + 8)n + 7 = nC0 + nC1 . 8 + nC2 . 82 + ....... + nCn 8n + 7. = 8. C1 + 82. C2 + ....... + Cn . 8n + 8. = 8, where  is a positive integer Hence, 9n + 7 is divisible by 8.

Solution :


8

MATHS Example # 12 : What is the remainder when 599 is divided by 13. Solution : 599 = 5.598 = 5. (25)49 = 5 (26 – 1)49 = 5 [49C0 (26)49 – 49C1 (26)48 + .......... + 49C48 (26)1 – 49C49 (26)0] = 5 [49C0 (26)49 – 49C1 (26)48 + ...........+ 49C48 (26)1 – 1] = 5 [49C0 (26)49 – 49C1(26)48 + .......... + 49C48 (26)1 – 13] + 60 = 13 (k) + 52 + 8 (where k is a positive integer) = 13 (k + 4) + 8 Hence, remainder is 8. Example # 13 : Find the last two digits of the number (17) 10. (17)10 = (289)5 = (290 – 1)5 = 5C0 (290)5 – 5C1 (290)4 + ........ + 5C4 (290)1 – 5C5 (290)0 = 5C0 (290)5 – 5C1 . (290)4 + .........5C3 (290)2 + 5 × 290 – 1 = A multiple of 1000 + 1449 Hence, last two digits are 49

Solution :

Note : We can also conclude that last three digits are 449. Example-14 : Which number is larger (1.01) 1000000 or 10,000 ? Solution :

By Binomial Theorem (1.01)1000000 = (1 + 0.01)1000000 = 1 + 1000000C1 (0.01) + other positive terms = 1 + 1000000 × 0.01 + other positive terms = 1 + 10000 + other positive terms Hence (1.01)1000000 > 10,000

Self practice problems : (8)

If n is positive integer, prove that the integral part of (5 5 + 11)2n + 1 is an even number..

(9)

If (7 + 4 3 )n =  + , where  is a positive integer and  is a proper fraction then prove that (1 – ) ( + ) = 1.

(10)

If n is a positive integer, then show that 32n + 1 + 2n + 2 is divisible by 7.

(11)

What is the remainder when 7103 is divided by 25 .

(12)

Find the last digit, last two digits and last three digits of the number (81) 25.

(13)

Which number is larger (1.2) 4000 or 800 Answers :

(11)

18

(12)

1, 01, 001

(13)

(1.2)4000.

Some standard expansions : (i)

Consider the expansion n

n

(x + y) = (ii)



r 0

n

Cr x n–r yr = nC x n y0 + nC x n–1 y1 + ...........+ nC x n–r yr + ..........+ nC x 0 yn ....(i) 0 1 r n

Now replace y  – y we get


9

MATHS n

n

(x – y) =



r 0

n

Cr (– 1) r x n–r yr = nC x n y0 – nC x n–1 y1 + ...+ nC (–1)r x n–r yr + ...+ nC (– 1)n x 0 yn ....(ii) 0 1 r n

(iii)

Adding (i) & (ii), we get (x + y)n + (x – y)n = 2[nC0 x n y0 + nC2 x n – 2 y2 +.........]

(iv)

Subtracting (ii) from (i), we get (x + y)n – (x – y)n = 2[nC1 x n – 1 y1 + nC3 x n – 3 y3 +.........]

Properties of binomial coefficients : (1 + x)n = C0 + C1x + C2x 2 + ......... + Cr x r + .......... + Cnx n where Cr denotes nCr (1)

......(1)

The sum of the binomial coefficients in the expansion of (1 + x) n is 2n Putting x = 1 in (1) n

C0 + nC1 + nC2 + ........+ nCn = 2n

......(2)

n

or (2)



n

Cr  2n

r 0

Again putting x = –1 in (1), we get n

C0 – nC1 + nC2 – nC3 + ............. + (–1) n nCn = 0

......(3)

n

or (3)

r n

Cr  0

r0

The sum of the binomial coefficients at odd position is equal to the sum of the binomial coefficients at even position and each is equal to 2n–1. from (2) and (3) n

(4)

 (1)

C0 + nC2 + nC4 + ................ = nC1 + nC3 + nC5 + ................ = 2n–1

Sum of two consecutive binomial coefficients n

Cr + nCr–1 =

L.H.S.

n+1

Cr

= nCr + nCr–1 =

n! n! + (n  r )! r! (n  r  1)! (r  1)!

n! 1  1 = (n  r )! (r  1)!     r n  r  1 =

n! (n  1) (n  r )! (r  1)! r(n  r  1)

(n  1)! = (n  r  1)! r! = (5)

n+1

Cr = R.H.S.

Ratio of two consecutive binomial coefficients


10

MATHS n n

(6)

n

Cr =

n r

Cr

n–1

nr 1 r

=

Cr 1

n(n  1) r(r  1)

Cr–1 =

n–2

n(n  1)(n  2).........(n  (r  1)) r (r  1)(r  2).......2 .1

Cr–2 = ............. =

Example # 15 : If (1 + x)n = C0 + C1x + C2x 2 + ............. + c nx n, then show that (i) C0 + 3C1 + 32C2 + .......... + 3n Cn = 4n.

Solution :

(ii)

C0 + 2C1 + 3. C2 + ........ + (n + 1) Cn = 2n – 1 (n + 2).

(iii)

C0 –

C2 C3 C1 Cn 1 + – + ......... + ( –1)n = . 3 2 4 n 1 n 1

(i) (1 + x)n = C0 + C1 x + C2x 2 + ........... + Cnx n put x = 3 C0 + 3 . C1 + 32 . C2 + .......... + 3n . Cn = 4n  Method : By Summation

(ii)

L.H.S. = nC0 + 2. nC1 + 3 . nC2 + ........ + (n + 1). nCn. n



n

 (r  1) .

=

n

Cr =

r 0

r 0

n



=n

n1

r 0

n

r. nCr +



n

Cr

r 0

n



Cr 1 +

r 0

n

Cr = n . 2n – 1 + 2n = 2n – 1 (n + 2). RHS

Method : By Differentiation (1 + x)n = C0 + C1x + C2x 2 + ........... + Cnx n Multiplying both sides by x, x(1 + x)n = C0x + C1x 2 + C2x 3 + ........ + Cn x n + 1. Differentiating both sides (1 + x)n + x n (1 + x)n – 1 = C0 + 2. C1x + 3 . C2x 2 + ....... + (n + 1)Cnx n. putting x = 1, we get C0 + 2.C1 + 3 . C2 + ...... + (n + 1) Cn = 2n + n . 2n – 1 C0 + 2.C1 + 3 . C2 + ...... + (n + 1) Cn = 2n – 1 (n + 2) Proved (iii)

 Method : By Summation C2 C3 C1 Cn + – + ........ + (– 1)n. 3 2 4 n 1

L.H.S. = C0 – n

=

 (1)

n

r

r 0

1 = n 1

Cr r 1

.

n

 (1)

r

r 0

=

1 [ n + 1C 1 – n 1

=

1 [– n 1

n+1

.

n +1

n+1

C0 +

C2 +

n+1

n  1 n . Cr   r 1

Cr + 1

n+1

C1 –

 Cr 1  

n 1

C3 – .............+ (– 1) n .

n+1

C2 + ......... + (– 1)n .

n+1

n+1

C n + 1]

Cn + 1 +

n+1


C 0]

11

MATHS 1 = R.H.S. , since n 1

=



n1

C0  n1 C1  n1 C2  ...  ( 1)n

n1



Cn1  0

Method : By Integration (1 + x)n = C0 + C1x + C2x 2 + ...... + Cn x n. Integrating both sides, within the limits – 1 to 0. 0

0

 (1  x )n  1   x2 x3 x n1   C2  .....  Cn   = C 0 x  C1  2 3 n  1   n  1  1 1 C1 C 2 C   1   .....  ( 1)n1 n  – 0 = 0 –  C 0  2 3 n  1  n 1

C0 –

C2 C1 Cn 1 + – .......... + (– 1) n = Proved 3 2 n 1 n 1

Example # 16 : If (1 + x)n = C0 + C1x + C2x 2 + ........+ Cnx n, then prove that (i) C02 + C12 + C22 + ...... + Cn2 = 2nCn (ii) C0C2 + C1C3 + C2C4 + .......... + Cn – 2 Cn = 2nCn – 2 or 2nCn + 2 (iii) 1. C02 + 3 . C12 + 5. C22 + ......... + (2n + 1) . Cn2 . = 2n. 2n – 1Cn + 2nCn. Solution : (i) (1 + x)n = C0 + C1x + C2x 2 + ......... + Cn x n. ........(i) n n n–1 n–2 0 (x + 1) = C0x + C1x + C2x + ....... + Cn x ........(ii) Multiplying (i) and (ii) (C0 + C1x + C2x 2 + ......... + Cnx n) (C0x n + C1x n – 1 + ......... + Cnx 0) = (1 + x)2n Comparing coefficient of xn, C02 + C12 + C22 + ........ + Cn2 = 2nCn (ii)

From the product of (i) and (ii) comparing coefficients of x n – 2 or x n + 2 both sides, C0C2 + C1C3 + C2C4 + ........ + Cn – 2 Cn = 2nCn – 2 or 2nCn + 2.

(iii)

 Method : By Summation L.H.S. = 1. C02 + 3. C12 + 5. C22 + .......... + (2n + 1) Cn2. n

=



r0

(2r  1) nC 2 = r

n

n

 2.r . ( C ) +  ( n

r 0

2

r

n

Cr )2

r0

n

=2

. n . r 1

n–1

Cr – 1 nCr + 2nCn

(1 + x)n = nC0 + nC1 x + nC2 x 2 + .............nCn x n ..........(i) n–1 n–1 n–1 n–1 n–2 n–1 0 (x + 1) = C0 x + C1 x + .........+ Cn – 1x .........(ii) Multiplying (i) and (ii) and comparing coeffcients of x n. n–1 C0 . nC1 + n – 1C1 . nC2 + ........... + n – 1Cn – 1 . nCn = 2n – 1Cn n



n1

Cr 1 . nCr = 2n – 1Cn

r 0

Hence, required summation is 2n. 2n – 1Cn + 2nCn = R.H.S.  Method : By Differentiation (1 + x 2)n = C0 + C1x 2 + C2x 4 + C3x 6 + ..............+ Cn x 2n Multiplying both sides by x x(1 + x 2)n = C0x + C1x 3 + C2x 5 + ............. + Cnx 2n + 1. Differentiating both sides


12

MATHS x . n (1 + x 2)n – 1 . 2x + (1 + x 2)n = C0 + 3. C1x 2 + 5. C2 x 4 + ....... + (2n + 1) Cn x 2n ........(i) (x 2 + 1)n = C0 x 2n + C1 x 2n – 2 + C2 x 2n – 4 + ......... + Cn ........(ii) Multiplying (i) & (ii) (C0 + 3C1x 2 + 5C2x 4 + ......... + (2n + 1) Cn x 2n) (C0 x 2n + C1x 2n – 2 + ........... + Cn) = 2n x 2 (1 + x 2)2n – 1 + (1 + x 2)2n comparing coefficient of x2n, C02 + 3C12 + 5C22 + .........+ (2n + 1) Cn2 = 2n . 2n – 1Cn – 1 + 2nCn. C02 + 3C12 + 5C22 + .........+ (2n + 1) Cn2 = 2n . 2n–1Cn + 2nCn. Proved Example # 17 : Find the summation of the following series – m (i) Cm + m+1Cm + m+2Cm + .............. + nCm n (ii) C3 + 2 . n+1C3 + 3. n+2C3 + ......... + n . 2n–1C3 Solution : (i) Method : Using property, nCr +nCr–1 = n+1Cr m Cm + m+1Cm + m+2Cm + .............. + nCm m 1 Cm1  m1 Cm + =   

m+2

Cm + .............. + nCm { mCm = m+1Cm+1}

m 2 Cm1  m 2 Cm  + .................. + nC =  m

= m+3Cm+1 + ............. + nCm = nCm+1 + nCm = n+1Cm+1  Method m Cm + m+1Cm + m+2Cm + .......... + nCm The above series can be obtained by writing the coefficient of x m in (1 + x)m + (1 + x)m+1 + ......... + (1 + x)n Let S = (1 + x)m + (1 + x)m+1 +.............. + (1 + x) n



(1  x )m 1  x 

=

n m 1

x

= coefficient of x m in (ii)



1

=

1  xn1 x

1  x n1  1  x m x –

1  xm x

=

n+1

Cm +1 + 0 =

n+1

Cm +1

n

C3 + 2 . n+1C3 + 3 . n+2C3 + .......... + n . 2n–1C3 The above series can be obatined by writing the coefficient of x 3 in (1 + x)n + 2 . (1 + x)n+1 + 3 . (1 + x)n+2 + ........... + n . (1 + x) 2n–1 Let S = (1 + x)n + 2 . (1 + x)n+1 + 3. (1 + x)n+2 + ........... + n (1 + x) 2n–1 (1 + x)S = (1 + x)n+1 + 2 (1 + x)n+2 + ............. + (n – 1) (1 + x) 2n–1 + n(1 + x)2n Subtracting (ii) from (i) – xS = (1 + x)n + (1 + x)n+1 + (1 + x)n+2 + .............. + (1 + x) 2n–1 – n(1 + x)2n =



 – n (1 + x)

(1  x )n (1  x )n  1

S=

x

 (1  x )2n  (1  x )n x

x3 : S x3 :

2

+

.....(i) ....(ii)

2n

n(1  x )2n x

(coefficient of x 3 in S)

 (1  x )2n  (1  x )n x2

+

n(1  x )2n x

Hence, required summation of the series is – 2nC5 + nC5 + n . 2nC4


13

MATHS Example # 18 : Prove that C1 – C3 + C5 – ........ = 2n/2 sin Solution :

or

n . 4

Consider the expansion (1 + x) n = C0 + C1 x + C2 x 2 + ...... + Cn x n putting x = – i in (i) we get (1 – i)n = C0 – C1i – C2 + C3i + C4 + ....... (– 1)n Cn in

....(i)

  n   n  cos    i sin    = (C – C + C – .......) – i (C – C + C – ...... ) ....(ii) 2n 2  0 2 4 1 3 5  4   4  Equating the imaginary part in (ii) we get C1 – C3 + C5 – ........ = 2n/2 sin

n . 4


Prove the following (i) C0 + 3C1 + 5C2 + ............. + (2n + 1) Cn = 2n (n + 1)

43 42 4n1 5n1  1 . C1 + C2 + .............. + Cn = 3 2 n 1 n 1

(ii)

4C0 +

(iii)

n

C0 . n+1Cn + nC1 . nCn–1 + nC2 .

(iv)

2

C2 + 3C2 + ......... + nC2 =

n+1

n–1

Cn–2 + ........... + nCn . 1C0 = 2n–1 (n + 2)

C3

Multinomial theorem : As we know the Binomial Theorem – n

n

(x + y)n =



n

Cr xn–r yr

 (n  r )! r!

=

n!

x n–r yr

r0

r 0

putting n – r = r1 , r = r2 therefore,

(x + y)n =



r1  r2

n! r ! r2 ! n 1

x r1 . y r2

Total number of terms in the expansion of (x + y) n is equal to number of non-negative integral solution of r1 + r2 = n i.e. n+2–1C2–1 = n+1C1 = n + 1 In the same fashion we can write the multinomial theorem (x 1 + x 2 + x 3 + ........... x k)n =



r1  r2 ...rk

n! x1r1 . x r22 ...x rkk r ! r !... r ! 1 2 k n

Here total number of terms in the expansion of (x 1 + x 2 + .......... + x k)n is equal to number of nonnegative integral solution of r 1 + r2 + ........ + rk = n i.e. n+k–1Ck–1

Example # 19 : Find the coefficient of a2 b3 c 4 d in the expansion of (a – b – c + d) 10

Solution :

(a – b – c + d)10 =



r1 r2  r3  r4

(10 )! r1 r2 r3 r4 r ! r ! r ! r ! (a) ( b) ( c ) (d) 10 1 2 3 4

we want to get a2 b3 c 4 d this implies that 

coeff. of a2 b3 c 4 d is

r 1 = 2, r2 = 3, r3 = 4, r4 = 1

(10 )! 3 4 2! 3! 4! 1! (–1) (–1) = – 12600


14

MATHS 11

7  Example # 20 : In the expansion of 1  x   , find the term independent of x. x 

Solution :

7  1  x   x 



11

=

r1  r2 r3

(11)! r ! r !r ! 11 1 2 3

r

73 (1)r1 ( x )r2   x

The exponent 11 is to be divided among the base variables 1, x and

7 in such a way so that we x

get x 0. Therefore, possible set of values of (r 1, r2, r3) are (11, 0, 0), (9, 1, 1), (7, 2, 2), (5, 3, 3), (3, 4, 4), (1, 5, 5) Hence the required term is

(11)! (11)! (11)! (11)! (11)! (11)! (70) + 9! 1 !1 ! 71 + 7! 2 ! 2 ! 72 + 5! 3 ! 3 ! 73 + 3! 4 ! 4 ! 74 + 1 ! 5 ! 5 ! 75 (11)! (11)! 2! (11)! 4! (11) ! 6! = 1 + 9 ! 2 ! . 1 ! 1 ! 71 + 7 ! 4 ! . 2 ! 2 ! 72 + 5 ! 6 ! . 3 ! 3 ! 73 (11) ! 8! (11) ! (10) ! + 3 ! 8 ! . 4 ! 4 ! 74 + 1 ! 10 ! . 5 ! 5 ! 75 = 1 + 11C2 . 2C1 . 71 + 11C4 . 4C2 . 72 + 11C6 . 6C3 . 73 + 11C8 . 8C4 . 74 + 11C10 . 10C5 . 75 5

=1+



11

r 1

C 2r . 2rCr . 7r


The number of terms in the expansion of (a + b + c + d + e + f) n is (A) n+4C4 (B) n+3Cn (C) n+5Cn

(16)

Find the coefficient of x 3 y4 z2 in the expansion of (2x – 3y + 4z) 9

(17)

Find the coefficient of x 4 in (1 + x – 2x 2)7

Answers :

(15)

C

(16)

9! 3 4 2 3! 4! 2! 2 3 4

(17)

(D) n + 1

– 91

Binomial theorem for negative and fractional indices : If n  R, then (1 + x)n = 1 + nx +

.................. +

n(n  1) n(n  1)(n  2) 2 x + x 3 + ................ 2! 3!

n(n  1)(n  2).......(n  r  1) x r + .................... . r!

Re mar ks (i)

The above expansion is valid for any rational number other than a whole number if | x | < 1.


15

MATHS (ii) (iii)

When the index is a negative integer or a fraction then number of terms in the expansion of (1 + x)n is infinite, and the symbol nCr cannot be used to denote the coefficient of the general term. The first term must be unity in the expansion, when index ‘n’ is a negative integer or fraction n 2     y y n (n  1)  y   x n 1    x n 1  n .     .....  if x x 2!  x      n (x + y) =   n 2    y n 1  x   y n 1  n . x  n (n  1)  x   .....  if         y y 2! y   

y 1 x

x 1 y

n(n  1)(n  2)......... (n  r  1) xr r!

(iv)

The general term in the expansion of (1 + x) n is T r+1 =

(v)

When ‘n’ is any rational number other than whole number then approximate value of (1 + x) n is 1 + nx (x 2 and higher powers of x can be neglected)

(vi)

Expansions to be remembered (|x| < 1) (a) (1 + x)–1 = 1 – x + x 2 – x 3 + .......... + (–1) r x r + ......... (b) (1 – x)–1 = 1 + x + x 2 + x 3 + .......... + x r + ......... (c) (1 + x)–2 = 1 – 2x + 3x 2 – 4x 3 + .......... + (–1)r (r + 1) x r + ........... (d) (1 – x)–2 = 1 + 2x + 3x 2 + 4x 3 + ............. + (r + 1)x r + ........... 

Example # 21 : Prove that the coefficient of x r in (1 – x)–n is Solution:

n+r–1

Cr

(r + 1)th term in the expansion of (1 – x) –n can be written as

n( n  1)( n  2)......( n  r  1) (–x)r r!

T r +1 =

= (–1)r

=

n(n  1)(n  2)......(n  r  1) n(n  1)(n  2)......(n  r  1) r (–x)r = x r! r!

(n  1)! n(n  1)......(n  r  1) r x (n  1) ! r ! (n  r  1)! n+r–1 = Cr Proved (n  1)! r!

Hence, coefficient of x r is

Example-22 : If x is so small such that its square and higher powers may be neglected, then find the value of

(1  3x )1/ 2  (1  x )5 / 3 ( 4  x )1/ 2

Solution :

(1  3x )1/ 2  (1  x )5 / 3 ( 4  x )1/ 2

=

3 5x x  1 1 / 2 x 1  2  19 x   2 3   1    = 1/ 2 6   4 2  x  21   4 

1 =

x 1  2  19 x  1  x  1  2  x  19 x  19 41     =   =1– – x =1– x 6 8 4 6 8     2  2  12 24


16

MATHS Self practice problems : (18)

Find the possible set of values of x for which expansion of (3 – 2x) 1/2 is valid in ascending powers of x.

(19)

If y =

(20)

The coefficient of x 100 in

2

3 1.3  2  2 1.3.5  2    + ............., then find the value of y2 + 2y + 2!   + 5 3! 5 5

(A) 100 Answers :

3  5x (1  x )2

is

(B) –57 (18)

 3 3 x   ,   2 2

(C) –197 (19)

4

(20)


(D) 53 C

17

Binomial Theorem

Recommend Documents