Question

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

• A. ${}^{1001}{C}_{50}$
• B. ${}^{1000}{C}_{50}$
• C. ${}^{1001}{C}_{51}$
• D. ${}^{1000}{C}_{51}$

Answer 1

The correct option is A ${}^{1001}{C}_{50}$

Clearly, the given series is a geometric progression in which
$a=(1+x{)}^{1000},r=\frac{x(1+x{)}^{999}}{(1+x{)}^{1000}}=\frac{x}{(1+x)}$ and$n=1001.$
$\therefore$ Given sum $=\frac{a(1-r^n)}{(1-r)}$
$=\frac{(1+x{)}^{1000}\times \{1-{\left(\frac{x}{1+x}\right)}^{1001}\}}{(1-\frac{x}{1+x})}$
$=(1+x{)}^{1001}-x^{1001}$

$\therefore$ Coefficient of $x^{50}$ in the above expansion $={}^{1001}{C}_{50}$