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

Answer: (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
$S=t^{1000}+x{t}^{999}+x^2{\left(t\right)}^{998}+...+x^{1000}$
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 ${}^{1001}{C}_{50}$
$=\frac{\left(1001\right)!}{50!\left(951\right)!}$