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 

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 $$= \binom{1002}{50}$$
