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This question already has an answer here:

So I recently found out about geometrical series. However now I've got this formula:

$\sum_{k=0}^{n-1} \frac 1 3 k \cdot {\frac 2 3}^k$ is there also a way to simplify this?

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marked as duplicate by Hans Lundmark, Daniel W. Farlow, астон вілла олоф мэллбэрг, user91500, E. Joseph Nov 18 '16 at 8:05

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    $\begingroup$ Hint: $\sum_k k\cdot q^k = q\cdot \sum_k k\cdot q^{k-1} = q\cdot \frac d{dq}\sum_k q^k$ $\endgroup$ – jameselmore Nov 16 '16 at 17:20
  • $\begingroup$ @jameselmore I don't understand your last step. Why is that equal and what is $d$? $\endgroup$ – fatalError Nov 16 '16 at 17:24
  • $\begingroup$ Magically multiply by $(1-q)^2$ and watch the cancellations. $\endgroup$ – Simply Beautiful Art Nov 16 '16 at 17:25
  • $\begingroup$ How did you do the geometric series formula? $\endgroup$ – Simply Beautiful Art Nov 16 '16 at 17:27
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    $\begingroup$ Do you know the finite sum-formula for a geometric series ? If yes, differentiate with respect to $q$ and look what happens. $\endgroup$ – Peter Nov 16 '16 at 17:44
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Let $S$ be the desired sum.

$$S=q+2q^2+3q^3+\dots+nq^n$$

Multiply it by $(1-q)$.

$$(1-q)S=(\color{red}q+\color{blue}{2q^2}+\color{red}{3q^3}+\dots+nq^n)-(\color{blue}{q^2}+\color{red}{2q^3}+\color{blue}{3q^4}+\dots+nq^{n+1})$$

Combine like terms. (notice the minus sign on the second set of values)

$$=\color{red}q+\color{blue}{[2-1]q^2}+\color{red}{[3-2]q^3}+\dots+[n-(n-1)]q^n-nq^{n+1}$$

$$=q(1+q+q^2+\dots+q^n)-nq^{n+1}$$

Let $G$ be the geometric sum.

$$(1-q)S=q(G-nq^n)$$

Solve for $S$.

$$S=\frac{q(G-nq^n)}{1-q}$$


The geometric sum:

$$G=1+q+q^2+\dots+q^n$$

Multiply it by $(1-q)$.

$$(1-q)G=(\color{red}1+\color{blue}q+\color{red}{q^2}+\dots+q^n)-(\color{blue}q+\color{red}{q^2}+\color{blue}{q^3}+\dots+q^{n+1})$$

Combine like terms. (notice the minus sign on the second set of values)

$$=\color{red}1+\color{blue}{[1-1]q}+\color{red}{[1-1]q^2}+\dots+[1-1]q^n-q^{n+1}$$

$$(1-q)G=1-q^{n+1}$$

Solve for $G$.

$$G=\frac{1-q^{n+1}}{1-q}$$

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  • $\begingroup$ I feel like a cheater now because someone posted me the right formula from Wikipedia. However I'd still like to understand it instead of just copy and and pasting it. But staring at this doesn't help me further. (edit: Please wait I saw that you added something maybe I get it now.) $\endgroup$ – fatalError Nov 16 '16 at 17:51
  • $\begingroup$ @fatalerror try adding the $q^2$ terms, then the $q^3$ terms, then the... And see what happens. $\endgroup$ – Simply Beautiful Art Nov 16 '16 at 17:53
  • $\begingroup$ I guess I will just go to sleep and look back at it tomorrow. I guess I'm just too tired to solve this right now... My brain is just barely working anymore... $\endgroup$ – fatalError Nov 16 '16 at 17:56
  • $\begingroup$ @fatalError Same. Got a headache to work with on my end. Anyways, I made everything more explicit. $\endgroup$ – Simply Beautiful Art Nov 16 '16 at 19:09

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