Question

The coefficient of $x^{50}$ in the binomial expansion of $(1 + x)^{1000} + x (1 + x)^{999} + x^2(1 + x)^{998} + .... + x^{1000}$ is:

• A. $\frac{(1000)!}{(50)! ( 950)!}$
• B. $\frac{(1000)!}{(49)! ( 951)!}$
• C. $\frac{(1001)!}{(51)! ( 950)!}$
• D. $\frac{(1001)!}{(50)! ( 951)!}$

Answer 1

The Correct Option is D

Let given expansion be $S = \left(1+x\right)^{1000} + x \left(1+x\right)^{999} + x^{2} \left(1+x\right)^{998} + ... + ...+ x^{1000}$ Put 1 + x = t $ S= t^{1000} + xt^{999} + x^{2} \left(t\right)^{998} + ...+x^{1000}$ This is a G.P with common ratio $ \frac{x}{t} $ $S = \frac{t^{1000} \left[ 1 - \left(\frac{x}{t}\right)^{1001}\right]}{1- \frac{x}{t}} $ $=\frac{\left(1+x\right)^{1000} \left[1- \left(\frac{x}{1+x}\right)^{1001}\right]}{ 1 - \frac{x}{1+x}}$ $ = \frac{\left(1+x\right)^{1001} \left[\left(1+x\right)^{1001} -x^{1001}\right]}{\left(1+x\right)^{1001}} $ $= \left[\left(1+x\right)^{1001} - x^{1001}\right]$ Now coeff of $x^{50}$ in above expansion is equal to coeff of $x^{50}$ in $(1 + x)^{1001}$ which is ${^{1001}C_{50}}$ $ = \frac{\left(1001\right)!}{50! \left(951\right)!} $