Mathematics - Binomial Theorem Question with Solution | TestHub

Let given expansion be $$\begin{array}{rl}& \mathrm{S}=(1+x{)}^{1000}+x(1+x{)}^{999}+x^2\\ & (1+x{)}^{998}+\dots +\dots +x^{1000}\end{array}$$ Put $1+x=t$ $$\mathrm{S}=t^{1000}+x{t}^{999}+x^2(t{)}^{998}+\dots +x^{1000}$$ This is a G.P with common ratio $\frac{x}{t}$ $$\begin{array}{rl}\mathrm{S}& =\frac{t^{1000}\left[1-{\left(\frac{x}{t}\right)}^{1001}\right]}{1-\frac{x}{t}}\\ & =\frac{(1+x{)}^{1000}\left[1-{\left(\frac{x}{1+x}\right)}^{1001}\right]}{1-\frac{x}{1+x}}\\ =& \frac{(1+x{)}^{1001}\left[(1+x{)}^{1001}-x^{1001}\right]}{(1+x{)}^{1001}}\end{array}$$ $$=\left[(1+x{)}^{1001}-x^{1001}\right]$$ Now coeff of $x^{50}$ in above expansion is equal to coeff of $x^{50}$ in $(1+x{)}^{1001}$ which is ${}^{1001}{C}_{50}=\frac{(1001)!}{50!(951)!}$