Coefficient of $x^{50}$ in the expansion Find the coefficient of $x^{50}$ in the expansion of $$(1+x)^{1000}+2x(1+x)^{999}+3x^2(1+x)^{998}+\cdots+1001x^{1000}$$
 A: Let $$S = (1 + x)^{1000} + 2x(1+x)^{999} +...+ 1000x^{999}(1+x)+ 1001 x^{1000}\tag1$$
This is an Arithmetic Geometric Series with $r = \frac{x}{1+x}$ and $d = 1$. Now $$\frac{x}{1+x}S = x(1 + x)^{999} + 2x^2(1 + x)^{998} +\cdots + 1000x^{1000} + \frac{1000x^{1001}}{1+x}\tag2$$
Subtracting we get,
$$(1 - \frac{x}{1+x}) S =(1+x)^{1000} + x(1+x)^{999} +\cdots + x^{1000} - \frac{1001x^{1000}}{1+x}$$
$$\Rightarrow S = (1+x)^{1001} + x(1+x)^{1000} + x^2(1+x)^{999} +...+ x^{1000}(1+x)-1001x^{1001}$$
This is a G.P. whose sum is $$S = (1+x)^{1002} - x^{1002} - 1002x^{1001}$$
So the coeff. of $x^{50}$ is $\binom{1002}{50}$. Hope it helps.                               
A: HINT:
Writing $n$ for $1000,$ we need $$S=\sum_{r=0}^n(r+1)(1+x)^{n-r}x^r$$
Now replace $n-r=m$ $$S=x^n\sum_{m=0}^n(n-m+1)\left(1+\dfrac1x\right)^m$$
$$=nx^n\sum_{m=0}^n\underbrace{\left(1+\dfrac1x\right)^m}_{\text{Binomial Expansion}}-x^n\sum_{m=0}^n(m-1)\left(1+\dfrac1x\right)^m$$
From wiki, $\sum_{k=1}^nkr^{k-1}=\dfrac{1-r^{n+1}}{(1-r)^2}-\dfrac{(n+1)r^n}{1-r}$
