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?

  • $\begingroup$ 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) $\endgroup$ – J.D. F. Mar 3 '20 at 15:52
  • $\begingroup$ No, multiplied sums cannot be split up $\endgroup$ – lioness99a Mar 3 '20 at 15:54
  • $\begingroup$ And to string together the two parts in your comment, remove the $$ from after the first part and from before the second $\endgroup$ – lioness99a Mar 3 '20 at 15:56
  • $\begingroup$ @J.D.F. The surest way is to use the partial sum of the geometric series. $\endgroup$ – callculus Mar 3 '20 at 16:58
  • 1
    $\begingroup$ Does this answer your question? Finding the sum of n terms $S_n$ starting from sigma $k=0$ $\endgroup$ – an4s Mar 3 '20 at 17:41

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


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.


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