What's the Summation formulae of the series $2*2^0 + 3*2^1 + 4*2^2 + 5*2^3.......$? I faced this question where I was asked to find a summation formulae for $n$ terms of $2*2^0 + 3*2^1 + 4*2^2 + 5*2^3.......$ I did try generalizing it with $$a_n = (n + 1)2^{n - 1}; n2^{n - 2}$$ but to no avail then I tried subtracting $2*(\sum 2^n)$ increasing geometric series from the above series( $2*2^0 + 3*2^1 + 4*2^2 + 5*2^3.......$) and I actually got something like a general term of $n2^n$ of the series obtained from this subtraction but then it lead me nowhere and I also think there is no scope for telescoping this kind of series.
I think the problem here is now I am devoid of any idea of how to approach this problem.

Let 
$$S_n(x)=\sum_{k=0}^n x^k=\frac{x^{n+1}-1}{x-1}.$$
Then
$$x\frac{dS_n(x)}{dx}=\sum_{k=0}^n kx^k=x\frac{(n+1)x^n}{x-1}-x\frac{x^{n+1}-1}{(x-1)^2}.$$
With $x=2$,
$$\sum_{k=0}^n 2^k=2^{n+1}-1,\\\sum_{k=0}^n k2^k=(n+1)2^{n+1}-2^{n+2}+2,$$
and 
$$\sum_{k=0}^n (2+k)2^k=2\left(2^{n+1}-1\right)+(n+1)2^{n+1}-2^{n+2}+2=(n+1)2^{n+1}.$$

Check:
$$2\cdot1+3\cdot2+4\cdot4+5\cdot8+6\cdot16\to2, 8, 24,64,160\to1\cdot2,2\cdot4,3\cdot8,4\cdot16,5\cdot32$$
