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