PROBLEM
Prove by induction $\sum_{i=1}^n i\cdot 2^{i-1} =(n-1)\cdot 2^n+1$
PROGRESS
So far I've proven $P_1$ and written $P_k$, but im stuck at $P_{k+1}$. I have written $P_{k+1}$ as
$$(k+1)\cdot 2^k+(k-1)\cdot 2^k+1=2^{k+1}\cdot k+1$$
but I don know how to simplify it, or even how to go on from here.
On
If you do not insist on looking for a induction method, I give another direct solution.
We want to evaluate $$S_n=1+2\cdot 2+3\cdot 2^2+\cdots+n \cdot 2^{n-1}.\tag1$$
But $$2\cdot S_n=2+2\cdot 2^2+3\cdot 2^3+\cdots+n \cdot 2^{n}.\tag2$$
Thus, by $(1)-(2)$, we have $$-S_n=1+2+2^2+2^3+\cdots+2^{n-1}-n \times 2^n=\frac{1(1-2^n)}{1-2}-n\cdot2^n=(1-n)2^n-1.\tag3$$
It follows that $$S_n=(n-1)2^n+1.$$
Define $S(k) = \sum\limits_{i=1}^k i\cdot 2^{i-1},$ then
\begin{aligned} S(k+1) = \sum\limits_{i=1}^{k+1} i \cdot 2^{i-1} &= (k+1)\cdot 2^{k} + \sum\limits_{i=1}^k i\cdot 2^{i-1}\\ &=(k+1)\cdot 2^k + S(k)\\ &= (k+1)\cdot2^k+(k-1)\cdot 2^k + 1\\ &= \big((k+1)+(k-1)\big)\cdot2^k+1\\ &= (2k)\cdot2^k + 1\\ &= k\cdot 2^{k+1}+1\\ \end{aligned} as desired. First, I split the sum into two parts: the final term and the rest. Then I identify that this rest is equal to $S(k)$ for which by assumption we know the value of. I insert the value of $S(k)$, then I factor out the $2^k$ multiplier, and then I do some final simplifications.