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

Question

The coefficient of $x^{50}$ in the binomial expansion of $(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\cdots+x^{1000}$ is:

Accepted Answer

This is a G.P. (in $1+x$) with ratio $\dfrac{x}{1+x}$ and 1001 terms, summing to
$$\frac{(1+x)^{1000}\left[1-\left(\frac{x}{1+x}\right)^{1001}\right]}{1-\frac{x}{1+x}}=(1+x)^{1001}-x^{1001}.$$

The coefficient of $x^{50}$ equals the coefficient of $x^{50}$ in $(1+x)^{1001}$, i.e. $\binom{1001}{50}=\dfrac{1001!}{50!\,951!}$.

Solutions · 1