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.

I can provide another idea:
$$
\begin{array}{cccccccc}
a= & 2\times2^{0}+ & 3\times2^{1}+ & 4\times2^{2}+ & 5\times2^{3}+ & \cdots & \left(M+1\right)\times2^{M-1}+ & \cdots\\
2a= &  & 2\times2^{1}+ & 3\times2^{2}+ & 4\times2^{3}+ & \cdots & \left(M\right)\times2^{M-1}+ & \cdots
\end{array}
$$
Then, by subtracting the first equation by the second one, you can get
$$
-a=2\times2^{0}+2+2^{2}+2^{3}+2^{4}+\cdots+2^{M-1}+\cdots
$$
If it is the finite sum, i.e., $a_M=\sum_{n=1}^{M}(n+1)2^{n-1}$, then the above equation has the last term which is negative, given by
$$
-a_{M}=2\times2^{0}+2+2^{2}+2^{3}+2^{4}+\cdots+2^{M-1}-\left(M+1\right)\times2^{M}
$$
I think you can get the result from here, which is $a_M = M\times2^M$.
