Question

The coefficient of $x^{50}$ in the expansion of $(1+x{)}^{1000}+2x(1+x{)}^{999}+3x^2(1+x{)}^{998}+......+1001x^{1000}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let, $S=(1+x{)}^{1000}+2x(1+x{)}^{999}+3x^2(1+x{)}^{998}+......+1001x^{1000}$
$\frac{x}{1+x}S=x(1+x{)}^{999}+2x^2(1+x{)}^{998}+........+1000x^{1000}+\frac{1001x^{1001}}{1+x}$
Substract above equations.
$(1-\frac{x}{1+x})S=(1+x{)}^{1000}+(1+x{)}^{999}+x^2(1+x{)}^{998}+....+x^{1000}-\frac{1001x^{1001}}{1+x}$
$\Rightarrow S=(1+x{)}^{1001}+x(1+x{)}^{1000}+x^2(1+x{)}^{999}+......+x^{1000}(1+x)-1001x^{1001}$
$=\frac{(1+x{)}^{1001}[{\left(\frac{x}{1+x}\right)}^{1001}-1]}{\frac{x}{1+x}-1}-1001x^{1001}$
[Sum of G.P]
$=(1+x{)}^{1002}-x^{1002}-1002x^{1001}$
$\therefore$ coefficient of $x^{50}$ in $S=$ coefficient of $x^{50}$ in $\left[\right(1+x{)}^{1002}-x^{1002}-1002x^{1001}]{=}^{1002}{C}_{50}$