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On Wikipedia the following closed form is derived - Generalised formula

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Can someone explain how the closed form below is derived?

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Solution thanks to graydad

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The first one is a pretty standard exercise. Let $S_k = \sum_{k=a}^b r^k$. $\space$Then $$rS_k = r\sum_{k=a}^b r^k = \sum_{k=a}^b r^{k+1}$$ so $$rS_k-S_k = (r^{a+1}+r^{a+2}+ \dots + r^{b+1})- (r^{a}+r^{a+1}+ \dots + r^{b}) \\ = r^{b+1}-r^{a}$$ and since $$rS_k-S_k = S_k(r-1)$$ we have enough to know that $$S_k(r-1)=r^{b+1}-r^{a} \\ \implies S_k = \frac{r^{b+1}-r^{a}}{r-1}$$ Try to use a similar tactic in deriving the second equality you are interested in, or just plug in $r = \frac{1}{\hat{r}}$ and solve in terms of $\hat{r}$ so that it looks like the second equation.

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  • $\begingroup$ Thanks, not sure how to turn k negative. $\endgroup$ – Chris Degnen Nov 26 '14 at 16:18
  • $\begingroup$ $r^{-k} = \frac{1}{r^k}$. Try making the substitution I mention in my last line. $\endgroup$ – graydad Nov 26 '14 at 16:19
  • $\begingroup$ Did you get it yet? $\endgroup$ – graydad Nov 26 '14 at 17:46
  • $\begingroup$ Yes, thanks. :-) $\endgroup$ – Chris Degnen Nov 26 '14 at 20:12
  • $\begingroup$ Great! Please accept my answer so that other users know this question is closed :) Or take the solution you added above and accept it as your own answer to this question. $\endgroup$ – graydad Nov 26 '14 at 20:16

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