If the coefficient of $x^{50}$ in the expansion of $(1+x)^{1000}+2x(1+x)^{999}+3x^2(1+x)^{998}$........ Problem : 
If the coefficient of $x^{50}$ in the expansion of $(1+x)^{1000}+2x(1+x)^{999}+3x^2(1+x)^{998} +\cdots +1001x^{1000}$ is $\lambda$  then the value of $\frac{1952! 50!}{1001!}\lambda$ 
Please guide how to find the value of $\lambda$ in this will be of great help as I am not getting any clue how to proceed further in this problem , Thanks 
 A: Let $$\displaystyle S = (1+x)^{1000}+2x(1+x)^{999}+..1000x^{999}(1+x)+1001x^{1000}........(1)\times \frac{x}{(1+x)}$$
So we get $$\frac{x\cdot S}{(1+x)} = x(1+x)^{999}+2x^2(1+x)^{998}+.........+1000x^{1000}+\frac{1001x^{1001}}{1+x}....(2)$$
So we get $$S\left[1-\frac{x}{1+x}\right] = (1+x)^{1000}+\underbrace{\left[x(1+x)^{999}+x^2(1+x)^{998}+......+x^{1000}\right]}_{S'}-\frac{1001x^{1001}}{1+x}$$
Now $$S'=x(1+x)^{999}+x^2(1+x)^{998}+.........+x^{1000}........(1)\times \frac{x}{1+x}$$
So $$\frac{S'\cdot x}{1+x} = x^2(1+x)^{998}+........+\frac{x^{1001}}{1+x}.......(2)$$
So we get $$\frac{S'}{1+x} = x(1+x)^{999}-\frac{x^{1001}}{1+x}\Rightarrow S'= x(1+x)^{1000}-x^{1001}$$
So we get $$\frac{S}{1+x}= (1+x)^{1000}+x(1+x)^{1000}-x^{1001}-\frac{1001x^{1001}}{1+x}$$
So we get $$S=(1+x)^{1002}-x^{1001}(1+x)-1001x^{1001}$$
Now Coefficient of $x^{50}$ in $$(1+x)^{1002}-x^{1001}(1+x)-1001x^{1001}$$ is $$= \binom{1002}{50}$$
A: You need to find the coefficients at $x^{50}$ in each term of the form $kx^{k-1}(1+x)^{1001-k}$, $k=1,\ldots,1001$.
By binomial expansion, the coefficients are $k \cdot \binom{1001-k}{50-(k-1)} = k   \binom{1001-k}{950}$. If $s<0$, then by convention we will put $\binom{n}{s}=0$.
Now we need to find 
$$\sum_{k=1}^{1001} k   \binom{1001-k}{950} = \sum_{k=1}^{51} k   \binom{1001-k}{950}=\sum_{s=0}^{50} (51-s)   \binom{950+s}{950},$$ and the latter sum should be easy to find.
A: $$\begin{align}
[x^s]\sum_{r=0}^n (r+1)x^r(1+x)^{n-r}&=[x^s]\sum_{r=0}^n(r+1)x^r\sum_{j=0}^{n-r}\binom {n-r}jx^j\\
&=[x^s]\sum_{r=0}^n (r+1) x^r\binom {n-r}{s-r}x^{s-r}\\
&=\sum_{r=0}^n \color{blue}{(r+1)} \binom {n-r}{s-r}\\
&=\sum_{r=0}^n \color{blue}{\sum_{u=0}^r}\binom {n-r}{n-s}\\
&=\sum_{u=0}^n\sum_{r=u}^n \binom {n-r}{n-s}
\color{lightgrey}{=\sum_{u=0}^n\sum_{r'=n-s}^{n-u}\binom {r'}{n-s}}\\
&=\sum_{u=0}^n\binom {n-u+1}{n-s+1}
\color{lightgrey}{=\sum_{u'=0}^{n}\binom {u'+1}{n-s+1}}\\
&=\binom {n+2}{n-s+2}=\binom {n+2}s\\
\text{Put $n=1000,s=50$ :}\qquad\qquad\qquad\\
[x^{50}]\sum_{r=0}^{1000}(r+1)x^r(1+x)^{1000-r}&=\binom {1002}{50}\quad\blacksquare\\
\end{align}$$

Alternatively: 
$$\begin{align}
&\sum_{r=0}^n(r+1)x^r(1+x)^{n-r}\\
&=\sum_{r=0}^n(r+1)x^r\sum_{j=0}^{n-r}\binom {n-r}jx^j\\
&=\sum_{r=0}^n\sum_{j=0}^{n-r}(r+1)\binom {n-r}{j}x^{r+j}\\
&=\sum_{r=0}^n\sum_{i=r}^n(r+1)\binom {n-r}{i-r}x^i &&(i=r+j)\\
&=\sum_{i=0}^nx^i\sum_{r=0}^i(r+1)\binom{n-r}{n-i} &&(0\le r\le i\le n)\\
&=\sum_{i=0}^nx^i\sum_{r=0}^i\sum_{u=0}^r\binom{n-r}{n-i} \\
&=\sum_{i=0}^nx^i\sum_{u=0}^i\sum_{r=u}^i\binom{n-r}{n-i} 
\color{lightgrey}{=\sum_{i=0}^nx^i\sum_{u=0}^i\sum_{r'=n-i}^{n-u}\binom{r'}{n-i} }&&(0\le u\le r\le i)\;  \color{lightgrey}{(r'=n-r)}\\
&=\sum_{i=0}^nx^i\sum_{u'=0}^i\binom {n-u+1}{n-i+1}
\color{lightgrey}{=\sum_{i=0}^nx^i\sum_{u'=n-i}^n\binom {u'+1}{n-i+1}}
&&\color{lightgrey}{(u'=n-u)}\\
&=\sum_{i=0}^n x^i\binom {n+2}{n-i+2}\\
&=\sum_{i=0}^n x^i\binom {n+2}i\\
\text{Put $n=1000$}&\text{ and select coefficient of $[x^{50}]$:}\\
&[x^{50}]\sum_{r=0}^{1000}(r+1)x^r(1+x)^{1000-r}\\
&=[x^{50}]\sum_{r=0}^{1000}x^i\binom {1002}i\\
&=\binom{1002}{50}\quad\blacksquare
\end{align}$$
