Closed form of $\sum_{i=0}^n ia^i$ derivation

Asked Jan 28 '18 at 00:17

Active Jan 28 '18 at 00:17

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$$\sum_{k=1}^n kz^k = z\frac{1-(n+1)z^n+nz^{n+1}+1}{(1-z)^2}$$

When I tried it on my own I only got as far as:

$$\sum_{i=0}^n ia^i = \frac{n(a^{n+1}-1)}{a-1}-a^n\sum_{i=0}^n(\frac{i}{a^i})$$

But then I realised that the second term is in fact $\sum_{i=0}^n (\frac{1}{a})^ii$ which is only the original problem with $z = \frac{1}{a}$ so I'm a bit stuck.

I should point out that this is a first year undergraduate maths course for engineering. That being said, I'd appreciate if you could point out what course might teach me how to solve this (or a good book).

asked Jan 28 '18 at 00:17

user1530491

1

Duplicate of Formula for calculating $\sum_{n=0}^{m}nr^n$, Formula for the summation of the series $a+2a^2+3a^3+...$ upto nth term? etc. – dxiv Jan 28 '18 at 00:23
Closed as duplicate. Thanks! – user1530491 Jan 28 '18 at 00:26
Not sure whether that was meant to be sarcastic, so I'll just note that having the question closed as a duplicate generally gives you the best, quickest answers. – dxiv Jan 28 '18 at 00:37
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Not at all! That was indeed what I was looking for! I appreciate your help, I was having a lot of trouble searching (what with variable names being different, I appreciate how the same thing could be asked over and over again). Thanks again! – user1530491 Jan 28 '18 at 01:42
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The site search leaves a lot to be desired math-wise, indeed. FWIW I located the duplicates using Approach0 as discussed here. – dxiv Jan 28 '18 at 01:51
Super useful! Thanks yet again! :D – user1530491 Jan 29 '18 at 00:25

Closed form of $\sum_{i=0}^n ia^i$ derivation

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