$$\sum_{i=0}^{n-1} ia^i = \frac{a-na^n+(n-1)a^{n+1}}{(1-a)^2}$$
What is the thought process behind obtaining this formula?
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1Welcome to the MSE, what have you tried? You could start by this answer. – Aaron Maroja Feb 10 '15 at 20:06
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I tried looking at it to no avail, the answer you've shown just shows another formula derived using this formula. – user214516 Feb 10 '15 at 20:14
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Try to write down what you have tried! Have you tried induction? – Aaron Maroja Feb 10 '15 at 20:15
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I don't want to prove this, I want to understand how someone figured it out. – user214516 Feb 10 '15 at 20:17
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The probably wrote down the cases for $n = 1,2,3,4,...$ and hypothesized an answer. – Simon S Feb 10 '15 at 20:28
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This question was asked and answered dozens of times. – Alex Feb 10 '15 at 20:30
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Could you provide the answer, searching wasn't succesful for me. – user214516 Feb 10 '15 at 20:38
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possible duplicate of How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$ – apnorton Feb 10 '15 at 21:31
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Well, here is my thought process. I let $S = \sum_{i = 0}^{n-1}ia^i$. Multiplying both sides by $a$, I get $aS = \sum_{i = 0}^{n-1}ia^{i+1}$. I want to express the sum on the right as a linear expression in $S$, for then I'll have a linear equation in $S$ for which I can solve. To this end, I write $$\sum_{i = 0}^{n-1}ia^{i+1} = \sum_{i = 0}^{n-1} [(i + 1)a^{i+1} - a^{i+1}] = \sum_{i = 0}^{n-1} (i + 1)a^{i+1} - \sum_{i = 0}^{n-1} a^{i+1}.$$ I notice that the expression $\sum_{i = 0}^{n-1} (i+1)a^{i+1}$ contains all the terms of $S$ plus one more term, $na^n$. So $\sum_{i = 0}^{n-1} (i + 1)a^{i+1} = S + na^n$. I know that the sum $\sum_{i = 0}^{n-1} a^{i+1}$ is a geometric sum with common ratio $a$ and first term $a$, so I have $$\sum_{i = 0}^{n-1} a^{i+1} = a\frac{a^n-1}{a-1} = \frac{a^{n+1}-a}{a-1}.$$ Collecting my results, I have $$aS = S + na^n - \frac{a^{n+1}-a}{a-1}.$$ That is, $$aS = S + \frac{na^n(a - 1) - a^{n+1} + a}{a - 1}.$$ The numerator on the second term on the right-hand side can be written out
$$na^{n+1} - na^n - a^{n+1} + a = (n-1)a^{n+1} - na^n + a.$$ Therefore,
\begin{align}aS &= S + \frac{(n-1)a^{n+1} - na^n + a}{a - 1}\\ (a - 1)S &= \frac{(n-1)a^{n+1} - na^n + a}{a - 1}\\ S &= \frac{(n-1)a^{n+1} - na^n + a}{(a - 1)^2}\\ S &= \frac{a - na^n + (n-1)a^{n+1}}{(1 - a)^2}. \end{align}
kobe
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@user214516: You can also prove the identity using derivatives and the function $f(x)= \sum_{i=1}^{n-1} x^i$. The identity is then an evaluation in $x=a$. The solution is more short than this one! I hope you can obtain the result by yourself using that. ;-) – rafaeldf Feb 10 '15 at 21:48
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@rafaeldf yes, although a longer approach, I decided to provide a method which does not use calculus. For that, this is what came to mind. – kobe Feb 10 '15 at 23:43