# HOW TO FIND THE CONSTANT TERM IN A BINOMIAL EXPANSION

Example 1 :

Find the constant term of (2x3 - (1/3x2))5

Solution :

=  (2x3 - (1/3x2))5

General term  Tr+1  =  nCr x(n-r) ar

x  = 2x3, n  =  5, a  =  (-1/3x2)

Tr+1  =  5Cr (2x3)5-r (-1/3x2)r

=  5Cr (2)5-r x15 - 3r (-1/3)x-2r

=  5Cr (-1/3)(2)5-r x15 - 5r

Constant term :

15 - 5r  =  0

15  =  5 r

r  =  15/5  =  3

5C3 (-1/3)(2)5-3 x15 - 5(3)

=  (-10/27) ⋅ 4  =  -40/27

So, the constant term is -40/27.

Example 2 :

Find the last two digits of the number 3600

Solution :

3600   =  (32) 300

=  (9)300

=  (10 - 1)300

=  300C0 10300 − 200C1 10299 + · · ·+ 300C298 102(−1)298 + 300C299 10(−1)299 + 300C300 (−1)300

= 102 [300C0 10298 − 200C1 10297 + · · ·+ 300C298 102(−1)298] - 3000 + 1

Multiple of 10 ends with 0. By subtracting 3000 from multiple of 10, we will get the value ends with 0.

Again by adding it by 1, we will get the value which ends with 01.

Example 3 :

If n is a positive integer, show that, 9n+1 − 8n − 9 is always divisible by 64.

Solution :

(1 + x)n+1  =  (n+1)C0 (n+1)C0 x + (n+1)Cx2 + (n+1)Cx3 + ......(n+1)Cn+1 xn+1

(1+8)n+1

=  (n+1)C(n+1)C18 + (n+1)C2(8)(n+1)C3(8)...... ....+  (n+1)Cn+1 8n+1

9n+1 − 8n − 9  =  [1 + (n+1)8 + 82 [(n+1)C2 (n+1)C38+(n+1)C482

...... ....+  (n+1)Cn+1 8n-1] - 8n - 9

9n+1 − 8n − 9  =  1 + 8n + 8 + [82 [(n+1)C2 (n+1)C38+(n+1)C482

...... ....+  (n+1)Cn+1 8n-1] - 8n - 9

9n+1 − 8n − 9  =  64 [An integer]

9n+1 − 8n − 9 is divisible by 64.

Example 4 :

If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)n are equal.

Solution :

If n is odd, then the two middle terms are T(n−1)/2+1 and T(n+1)/2+1

General term  Tr+1  =  nCr x(n-r) ar

x  =  x, n  =  n, a  =  y and r = (n−1)/2

nCx  =  T(n−1)/2+1  =  nC(n−1)/2 x(n-r) y(n−1)/ ----(1)

nCy  =  T(n+1)/2+1  =  nC(n+1)/2 x(n-r) y(n+1)/2  ----(2)

If nCx  =  nC ==> then x + y = n

Evidently if x + y = n then nCx  =  nCy

=  (n-1)/2 + (n + 1)/2

=  2n/2

=  n

So, the coefficients of middle terms are equal.

Example 5 :

If n is a positive integer and r is a non negative integer, prove that the coefficients of xr and xn−r in the expansion of (1 + x)n are equal

Solution :

General term  Tr+1  =  nCr x(n-r) ar

x = 1, a = x, n = n

 Tr+1  =  nCr 1(n-r) xr =  nCr xr  ----(1) Tr+1  =  nCn-r 1(n-(n-r)) x(n-r) =  nCn-r x(n-r) ----(2)

By comparing the coefficients of (1) and (2)

nC =  nCn-r

So, they are equal.

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