Binomial theorem notes

BINOMIAL THEOREM
MAIN CONCEPTS AND RESULTS
** Binomial theorem for any positive integer n
n
a  bn  n C0a n  n C1a n 1.b  n C2a n  2 .b2  ... n Cn 1a.bn 1  n Cn bn   n Ck a n  k bk
k 0
n
** The coefficients C r occurring in the binomial theorem are known as binomial coefficients.
** There are (n +1) terms in the expansion of a  b  , i.e., one more than the index.
n


** 1  x   n C0  n C1x  n C 2 x 2  ...  n C n 1x n 1  n C n x n
n


** 1  x   n C0  n C1x  n C 2 x 2  ...   1 n C n x n .
n n


** n C0  n C1  n C2  n C3  ...  n Cn  2n .
** n C0  n C1  n C2  n C3  ...   1 n Cn  0 .
n 1


** General Term in the expansion of a  b  = t r 1 n Cra n r .br
n


** Middle term in the expansion of a  b 
n

th th th
n   n 1  n 1 
(i)   1 term if n is even (ii)   and   1 terms if n is odd.
2   2   2 
**Pascal’s Triangle




ILLUSTRATIONS :

EXAMPLE 1:No. of terms in 1 + 3x + 3x 2 + x 3 6
is

(A) 7 (B) 6 (C) 17 (D) 19
Ans: D 19
1 + 3x + 3x 2 + x 3 6 = 1 + x)3 6
= (1 + x)18
n=18 ∴ no. of terms = 19

EXAMPLE 2:nC 0 + nC 1 + nC 2 + ⋯ + nC n −1 + nC n =

(A)2n (B) 2n (C) 2n−1 (D) none of these

Ans: B 2n
(1 + a)n = nc 0 a0 + nc 1 a1 + nc 2 a2 + ⋯ + nc n −1 an−1 + nc n an
put a=1

EXAMPLE 3:In 1 + x3 5
coefficient of x 9 is

51

, (A) 5 (B) 10 (C) 1 (D) NONE
Ans: B 10
1 + x 3 5 = 5C 0 + 5C 1 x 3 + 5C 2 (x 3 )2 + 5C 3 (x 3 )3 + 5C 4 (x 3 )4 + 5C 5 (x 3 )5

= 5C 0 + 5C 1 x 3 + 5C 2 x 6 + 5C 3 x 9 + 5C 4 x12 + 5C 5 x15

∴the coefficient of x 9 is 5C 3 i.e. 10
9 9
EXAMPLE 4:Number of terms in 1 + 5 2x + 1 − 5 2x ifx > 0 𝑖𝑠 _______

(A) 3 (B) 5 (C) 4 (D) 6
Ans: B 5

2 8
EXAMPLE 5:Find 6th term in 3x −
3x

Solution :Tr+1 = nC r x n−r ar

2 5
T6 = T5+1 = 8C 5 (3x)8−5 (− )
3x
2 5 1729 1
=56 ×33 ×x 3 × − 3x = - 9 x2

EXAMPLE 6:Find the value of (101)4 by using Binomial theorem :
n
Solution : 1 + x = nC 0 x 0 + nC 1 x1 + nC 2 x 2 + ⋯ + nC n −1 x n−1 + nC n x n
4 4
101 = 1 + 100

= 4C 0 (100)0 + 4C 1 (100)1 + 4C 2 (100)2 + 4C 3 (100)3 + 4 100 4


= 1 + 4 100 + 6 10000 + 4 1000000 + 100000000 = 104060401

EXAMPLE 7:Which of the following is larger9950 + 10050 or10150
50 50
Solution : 101 = 1 + 100

= 50C 0 (100)50 + 50C 1 (100)49 + 50C 2 (100)48 + ⋯ + 50C 49 (100)1 + 50C 50 100 0


50 50
99 = 100 − 1

= 50C 0 (100)50 − 50C 1 100 49
+ 50C 2 100 48
+ ⋯ − 50C 49 (100)1 + 50C 50 100 0


50 50
101 − 99 =2{50C 1 (100)49 + 50C 3 (100)47 + ⋯ + 50C 49 (100)1 }
50 50
101 − 99 =100 × (100)49 + 2{50C 3 (100)47 + ⋯ + 50C 49 (100)1 }

50 50
101 − 99 − (100)50 =2 50C 3 100 47
+ 50C 4 100 46
+ ⋯ + 50C 49 100 1
>0
50 50
101 > 99 + (100)50
50
Therefore , 101 islarger




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