Question

Find the coefficient of $x^{50}$ in the expression $(1+x{)}^{1000}+2x(1+x{)}^{999}+3x^2(1+x{)}^{998}+\dots \dots \dots +1001x^{1000}$

Answer 1

Coefficient of $x^{50}$ in

$S=(1+x{)}^{1000}+2x(1+x{)}^{999}+3x^2(1+x{)}^{998}+....+10001x^{1000}$

Multiply by $\frac{{}^{\prime \prime }{x}^{\prime \prime }}{1+x}$

$\frac{xS}{(1+x)}=x(1+x{)}^{999}+2x^2(1+x{)}^{998}+.....+\frac{1001x^{1001}}{1+x}$

$s[1-\frac{x}{1+x}]=(1+x{)}^{1000}+\left[x\right(1+x{)}^{999}+x^2(1+x{)}^{998}+.....+x^{1000}]\frac{-1001x^{1001}}{1+x}$

Now

$s^{\prime }=x(1+x{)}^{999}+x^2(1+x{)}^{998}+....+x^{1000}$

Multiply by $\frac{x}{1+x}$

$\Rightarrow \frac{S^{\prime }}{1+x}=x^2(1+x{)}^{998}+.....+\frac{x^{1001}}{1+x}$

$\Rightarrow \frac{S^{\prime }}{1+x}=x(1+x{)}^{998}-\frac{x^{1001}}{1+x}$

$S^{\prime }=x(1+x{)}^{1000}-x^{1001}$

$\Rightarrow \frac{s}{1+x}=(1+x{)}^{1000}+x(1+x{)}^{1000}-x^{1001}-\frac{1001x^{1001}}{1+x}$

$s=(1+x{)}^{1002}-x^{1001}(1+x)-1001x^{1001}$

Coefficient of $x^{50}$ in

$(1+x{)}^{1002}-x^{1001}(1+x)-1001x^{1001}$

${=}^{1001}{C}_{50}$.