Question

The coefficient of $x^{50}$ in the expansion of $(1+x{)}^{1000}+x(1+x{)}^{999}+x^2(1+x{)}^{999}+\cdots +x^{1000}$ is ${}^n{C}_k.$ Then the least value of $n+k$ is equal to

• A. $1050$
• B. $1049$
• C. $1051$
• D. $1005$

Answer 1

The correct option is C $1051$

Let $y=(1+x{)}^{1000}+x(1+x{)}^{999}+x^2(1+x{)}^{999}+\cdots +x^{1000}$
For the above GP
$a=(1+x{)}^{1000},$ $r=\frac{x}{1+x},$ $n=1001$
$y=\frac{(1+x{)}^{1000}[1-{\left(\frac{x}{1+x}\right)}^{1001}]}{1-\left(\frac{x}{1+x}\right)}=\frac{(1+x{)}^{1000}-\frac{x^{1001}}{1+x}}{\frac{1}{1+x}}=(1+x{)}^{1001}-x^{1001}$
Coefficient of $x^{50}={}^{1001}{C}_{50}$
$\therefore n+k=1051$