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My Precalc teacher gave me this as a question and I simply cannot figure out how to do it.

user26857
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user237616
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2 Answers2

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Let's call this sum $S_n(x)$, i.e. $$ S_n(x)=\sum_{k=0}^{n-1}e^{kx}. $$ Clearly $S_n(0)=n$. For $x\ne 0$, if you multiply $S_n(x)$ by $e^x$, you get: \begin{eqnarray} e^xS_n(x)&=&e^x\sum_{k=0}^{n-1}e^{kx}=\sum_{k=0}^{n-1}e^{(k+1)x}=e^x+e^{2x}+\ldots+e^{(n-1)x}+e^{nx}\\ &=&-\color{blue}{1}+\underbrace{[\color{blue}{1}+e^x+e^{2x}+\ldots+e^{(n-1)x}]}_{S_n(x)}+e^{nx}\\ &=&-1+S_n(x)+e^{nx}. \end{eqnarray} Solving $$ e^xS_n(x)=-1+S_n(x)+e^{nx} $$ for $S_n(x)$ you get: $$ S_n(x)=\frac{1-e^{nx}}{1-e^x}. $$ Hence $$ \sum_{k=0}^{n-1}e^{kx}=\begin{cases} n &\mbox{ for } x=0\\ \frac{1-e^{nx}}{1-e^x} &\mbox{ for } x\ne 0 \end{cases}. $$

HorizonsMaths
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Use the identity $$ 1+a+a^2+\dotsb+a^{n-1}=\frac{a^n-1}{a-1} $$ Because if $S=1+a+a^2+\dotsb+a^{n-1}$ then $$ aS=a+a^2+\dotsb+a^{n}=S+a^n-1 $$

k1.M
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  • What could i do if I forget this identity during a test? – user237616 May 03 '15 at 23:50
  • Take a look at new edit. – k1.M May 03 '15 at 23:55
  • What could i do if I forget this identity during a test? Figuring it out today by yourself (instead of having us do it for you) would have helped you remember it... – GEdgar May 04 '15 at 00:05
  • Well to figure it out myself, I need to understand how one derives it to actually apply it – user237616 May 04 '15 at 01:23
  • @user237616 If you forget it during a test and cannot remember how to derive it, then all you can do is write that you understand this is a geometric series, that $r = e^x$, and that there is a formula to find the sum of such a series but you cannot remember it. A good teacher should give some partial credit for that (a bad teacher won't). – Jared May 05 '15 at 23:21