Stuck in this sum using the Perturb the Sum Method.

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?

• Can I transform $$\sum_{k=1}^n k(2^k)$$ into $$\sum_{k=1}^n (2^k)\sum_{i=1}^k 1$$ ? (2nd and 3rd sums should be stringed together) – J.D. F. Mar 3 '20 at 15:52
• – lioness99a Mar 3 '20 at 15:54
• And to string together the two parts in your comment, remove the  from after the first part and from before the second – lioness99a Mar 3 '20 at 15:56
• @J.D.F. The surest way is to use the partial sum of the geometric series. – callculus Mar 3 '20 at 16:58
• Does this answer your question? Finding the sum of n terms $S_n$ starting from sigma $k=0$ – an4s Mar 3 '20 at 17:41

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