Summation involving the index multiplied by an exponential?

Question

$$\begin{align} \sum_{i=0}^{n-1}ia^i &= 0 + a + 2a^2 + \cdots + (n-1)a^{n-1}\\ &= \frac{a - na^n + (n-1)a^{n+1}}{(1 - a)^2} \end{align}$$

Is there any formula for a slightly similar one?

$$ \sum_{i=0}^{n-1}ia^{n-1-i} = 0 + a^{n-2} + 2a^{n-3} + \cdots + (n-1) $$

As Calvin Lin suggestion:

$$\begin{align} \sum_{i=0}^{n-1}ia^{n-1-i} &= \sum_{i=0}^{n-1}ia^{n-1}\left(\frac{1}{a}\right)^i = a^{n-1}\sum_{i=0}^{n-1}i\left(\frac{1}{a}\right)^i \\ &= a^{n-1}\frac{\frac1a - n\left(\frac1a\right)^n + (n-1)\left(\frac1a\right)^{n+1}}{(1 - \frac1a)^2}\\ &= \cdots \end{align}$$

SO, the final simplification will become $ \frac {a^n - an + n - 1} { (a-1)^2}$. — Calvin Lin, Mar 21 '13 at 21:38

score 3 · Accepted Answer · answered Mar 21 '13 at 16:39

3

Hint: $a^{n-1 - i} = a^{n-1} \times (\frac {1}{a} )^i$.

Now apply your initial claim.

answered Mar 21 '13 at 16:39

Calvin Lin

68,864

Thanks, see my edit... am I right? – user34295 Mar 21 '13 at 16:45
@Gremo Yes, and do you know how to complete it? – Calvin Lin Mar 21 '13 at 20:44
I think I should use the first formula with $a = \frac1a$, right? – user34295 Mar 21 '13 at 20:45
@Gremo Correct. Can you fill in what the final term would end up being? – Calvin Lin Mar 21 '13 at 20:46
Ok, it should work, see my edits. It may be simplified I think. – user34295 Mar 21 '13 at 21:09
@Gremo You can simplify it further by multiplying the $a^{n-1}$ in, which greatly simplifies your formula. Also, your initial claim is wrong, and it shouldn't involve an $i$, and it should be a 1 instead. – Calvin Lin Mar 21 '13 at 21:16

Summation involving the index multiplied by an exponential?

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