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I have a summation that I calculated, but I'm having a hard time to turn it into a geometric series.

I have the summation:

$y(n) = \sum_{k=-\infty}^{\infty} k*a^ku(k)*u(n-k)$ where $u(k)$ is the step function.

I reduced this to:

$y(n) = \sum_{k=0}^{n} k*a^k$ where $n >= 0$.

Since this is a finite series, I was taught that I could find a ratio of the second to first term, and use the following equation:

$y(n) = (firstterm) * {1-R^{n+1}\over1-R}$ Where "R" is the ratio of first term to second term.

However, I can't figure out what "R" should be since the first term is zero. How can I go about calculating "R" and $y(n)$?

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Not every finite series is a geometric series. There is no $R$ for your series, and the formula you are trying to apply, doesn't apply. –  Gerry Myerson Mar 1 '13 at 1:33
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2 Answers

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Let $$y(n)=a+2a^2+3a^3+\cdots+na^n$$ Then $$ay=a^2+2a^3+\cdots+na^{n+1}$$ Subtracting, $$(1-a)y=a+a^2+\cdots+a^n-na^{n+1}$$ Now the first $n$ terms do form a geometric series, so $$(1-a)y={a(1-a^n)\over1-a}-na^{n+1}$$ Now you can divide both sides by $1-a$ (provided $a\ne0$) to get a formula for $y$.

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This is really helpful! Now I understand your comment. –  Chris Harris Mar 1 '13 at 1:41
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And, by induction, once you have a formula for $\sum_{j=0}^n a^j j^k$, you can get a formula for $\sum_{j=0}^n a^j j^{k+1}$.

Gerry Myerson showed how to go from $k=0$ to $k=1$.

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