Mathematics - Binomial Theorem Question with Solution | TestHub

Let, $S=(1+x{)}^{1000}+2x(1+x{)}^{999}+2x(1+x{)}^{999}+3x^2(1+x{)}^{998}+\dots ..+1001x^{1000}$ $\frac{x}{1+x}S=x(1+x{)}^{999}++2x^2(1+x{)}^{998}+\dots \dots$ $+1000_x{}^{1000}+\frac{1001x^{1001}}{1+x}$ Subtract above equations, $\left(1-\frac{x}{1+x}\right)S=(1+x{)}^{1000}+(1+x{)}^{999}+$ $x^2(1+x{)}^{998}+\dots +x^{1000}-\frac{1001x^{1001}}{1+x}$ $\Rightarrow S=(1+x{)}^{1001}+x(1+x{)}^{1000}+x^2(1+x{)}^{999}$ $+\dots \dots \dots +x^{1000}(1+x)-100]x^{1001}$ $=\frac{(1+x{)}^{1001}\left[{\left(\frac{x}{1+x}\right)}^{1001}-1\right]}{\frac{x}{1+x}-1}-1001x^{1001}$ [0pt] [sum of G.P] $=(1+x{)}^{1002}-x^{1002}-1002x^{1001}$ ∴ coefficient of $x^{50}$ in $\mathrm{S}=$ coefficient of $x^{50}$ in $\left[(1+x{)}^{1002}-x^{1002}-1002x^{1001}\right]={}^{1002}{C}_{50}$