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I need to find the summation of $ab^{-k}$ from $k=5$ to $n$ using Gauss' Law. Here's what I have so far:

$$\begin{align}S_n&=(ab^{-5}+ab^{-6}+ab^{-7}+\cdots+ab^{-n}+ab^{-n}+ab^{-(n-1) }+ab^{-(n-2) } +\cdots+ab^{-5})\\2S&=(\frac{a}{b^5} +\frac{a}{b^n})+(\frac{a}{b^6}+\frac a{b^{-n+1}} )+(\frac a{b^n} +\frac{a}{b^{-n+2}} )+\cdots+(\frac{a}{b^n} +\frac{a}{b^5} ) \\ 2S&=(\frac a{b^5}) (1+\frac 1 b+\frac 1 {b^2} +\cdots+\frac{1}{b^{n-4}})(n-4)\end{align}$$

From here I'm confused how to get the answer. What should my next step be?

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$$ab^{-5}+ab^{-6}+ab^{-7}+\cdots+ab^{-n}=ab^{-n}\left(1+b+b^2+\cdots+b^{n-5}\right)$$ $$ab^{-5}+ab^{-6}+ab^{-7}+\cdots+ab^{-n}=\dfrac{ab^{-n}}{b-1}\left(b^{n-4}-1\right),\,\,\,\,\,b\not=1.$$ $$ab^{-5}+ab^{-6}+ab^{-7}+\cdots+ab^{-n}=\dfrac{a}{b-1}\left(\dfrac{1}{b^4}-\dfrac{1}{b^n}\right)$$

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  • $\begingroup$ Wolfram alpha has the answer as $$\begin{align}(ab^{-n-4}(b^n-b^4))/(b-1) \end{align}$$ which is the same except for the missing (b^(-n-4)) term? $\endgroup$ – dms94 Mar 29 '15 at 19:24
  • $\begingroup$ Both answers are same. You have just simplify my one to obtain the WA's answer. Good Luck. $\endgroup$ – Bumblebee Mar 30 '15 at 10:38

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