Question

If a and d are two complex numbers, then the sum to $x^{3}$ terms of the following series

$(1 + x)^{m}(1 - x)^{n}$is.

• A. $x^{m}$
• B. $\frac{(2n)!}{(m)!(2n - m)!}$
• C. 0
• D. None of these

Answer 1

Answer: (C)

0

Answer 2

We can write

$C_{0}C_{2} + C_{1}C_{3} + C_{2}C_{4} + C_{n - 2}C_{n}$upto $\frac{(2n)!}{(n + 1)!(n + 2)!}$terms

$\frac{(2n)!}{(n - 2)!(n + 2)!}$ ....(i)

Again,$\frac{(2n)!}{(n)!(n + 2)!}$ ...(ii)

Differentiating with respect to x,

$\frac{(2n)!}{(n - 1)!(n + 2)!}$ ....(iii)

Putting x =1 in (ii) and (iii), we get

$(1 + x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + ... + C_{n}x^{n}$

and $C_{0} + C_{2} + C_{4} + C_{6} + .....$

Thus the required sum to (n+1) terms, by (i)

= a.0 + d.0 = 0.