The coefficient of x^50 in the binomial expansion of (1+x)^1000+x(1+x)^999+x^2(1+x)^998+… +x^1000 is:

Let given expansion be $$\begin{aligned} &\mathrm{S}=(1+x)^{1000}+x(1+x)^{999}+x^2 \\ &(1+x)^{998}+\ldots+\ldots+x^{1000} \end{aligned}$$​S=(1+x)1000+x(1+x)999+x2(1+x)998+…+…+x1000​ Put $1+x=t$1+x=t $$\mathrm{S}=t^{1000}+x t^{999}+x^2(t)^{998}+\ldots+x^{1000}$$S=t1000+xt999+x2(t)998+…+x1000 This is a G.P with common ratio $\frac{x}{t}$tx​ $$\begin{aligned} \mathrm{S} &=\frac{t^{1000}\left[1-\left(\frac{x}{t}\right)^{1001}\right]}{1-\frac{x}{t}} \\ &=\frac{(1+x)^{1000}\left[1-\left(\frac{x}{1+x}\right)^{1001}\right]}{1-\frac{x}{1+x}} \\ =& \frac{(1+x)^{1001}\left[(1+x)^{1001}-x^{1001}\right]}{(1+x)^{1001}} \end{aligned}$$S=​=1−tx​t1000[1−(tx​)1001]​=1−1+xx​(1+x)1000[1−(1+xx​)1001]​(1+x)1001(1+x)1001[(1+x)1001−x1001]​​ $$=\left[(1+x)^{1001}-x^{1001}\right]$$=[(1+x)1001−x1001] Now coeff of $x^{50}$x50 in above expansion is equal to coeff of $x^{50}$x50 in $(1+x)^{1001}$(1+x)1001 which is ${ }^{1001} C_{50}=\frac{(1001) !}{50 !(951) !}$1001C50​=50!(951)!(1001)!​