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I have the following summation which I need to find the partial sum formula for:

$$ \sum_{i=1}^n ia^i $$

I tried the following manipulation:

$$\begin{align} S_n & = a + 2a^2 + 3a^3 + \cdots + na^n \\ aS_ n & = a^2 + 2a^3 + 3a^4 + \cdots + na^{n+1} \\ S_n - aS_n & = a - na^{n+1} \\ S_n (1 - a) & = a - na^{n+1} \\ S_n & = \frac {a - na^{n+1}}{1 - a} = \frac {a (1 - na^{n})} {1 - a} \end{align}$$

But WolframAlpha says this is incorrect. Could someone explain where I went wrong? Am I using the correct method for this type of problem?

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  • $\begingroup$ Hint: $S_n - aS_n = a + a^2 + a^3 + a^4 + \ldots + a^n - na^{n+1}$. $\endgroup$ – Jose Arnaldo Bebita-Dris Sep 15 '14 at 12:48
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$S_n-aS_n=a+a^2+a^3+\cdots+a^n-na^{n+1}$

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According to https://www.tug.org/texshowcase/cheat.pdf $$\sum_{i=0}^n ia^i = \frac{na^{n+2}-(n+1)a^{n+1}+a}{(a-1)^2}, \quad a \neq 1$$ $$\sum_{i=0}^n ia^i = \frac{a}{(a-1)^2}, \quad |a|<1$$

Note that, $$S_n-aS_n = a+a^2+a^3+\cdots+a^n-na^{n+1}$$ Now, the series on righthand side is a geometric series with ratio $r=a$. $$S_n(1-a) = \frac{a(1-a^n)}{(1-a)}-na^{n+1}$$ $$S_n = \frac{a(1-a^n)}{(1-a)^2}-\frac{na^{n+1}(1-a)}{(1-a)^2}$$ $$ = \frac{a-a^{n+1}-na^{n+1}+na^{n+2}}{(1-a)^2}$$ $$ = \frac{a-(1+n)a^{n+1}+na^{n+2}}{(1-a)^2}$$

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    $\begingroup$ what a cool cheatsheet that is! $\endgroup$ – hypergeometric Sep 15 '14 at 14:34
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Note that $$S_n-aS_n\not =a-na^{n+1}$$ and that $$S_n-aS_n=a+a^2+a^3+\cdots+a^n-na^{n+1}.$$

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