Stuck in this sum using the Perturb the Sum Method.

Question

The sum appears to be simple, but I must have dedicated over 3 hours to solve this and I just can't seem to do it

$$\sum_{k=0}^n k(2^k)$$

To solve using the perturb the sum method.

This was my latest attempt:

$$Sn = \sum_{k=0}^n k(2^k) = \sum_{k=1}^n (2^k)\sum_{i=1}^k 1 $$ Into:

$$ S(n+1) = 2 + \sum_{k=2}^{n+2} (2^k) \sum_{i=2}^k 1 = \sum_{k=1}^{n} (2^k) \sum_{i=1}^k 1 + 2^{n+1}(n+1)$$

I tried to transform the left side sums into the original ones but multiplying by 2 and removing the sum of 2^k and I got close to the answer, but it was still wrong. From what has been said, I reckon the very first step into transforming the original sum into two others is wrong, right?

Answer 1

Hint: First you start with the partial sum of a geometric series

$$\sum_{k=0}^nx^k=\frac{x^{n+1}-1}{x-1}$$

Next you differentiate both sides w.r.t $x$ $$\sum_{k=0}^n kx^{k-1}=\frac{d}{dx}\frac{x^{n+1}-1}{x-1}$$ $$\sum_{k=0}^n kx^{k}=x\frac{d}{dx}\frac{x^{n+1}-1}{x-1}$$

At last you calculate $\frac{d}{dx}\frac{x^{n+1}-1}{x-1}$ by using the quotient rule.