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This is my first time on here. I am in first year engineering, and I'm having some trouble with sigma notation. Here is the question and answer: enter image description here

I am trying to convert the summation into closed form, using the rules of summation, and the summation formulas. I was able to get the same answer, except that my answer did not multiply the first term by pi. Is there anyone able to explain how to do this, or give me some insight so that I might try it again in a different way? Thank you so much. Your time is greatly appreciated.

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  • $\begingroup$ Note that the summation starts from $k=1$ rather than $k=0$ which is what you evaluated it for. $\endgroup$ – Ali Caglayan Jan 13 '15 at 23:53
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    $\begingroup$ Thanks a lot guys. I did eventually figure it out. I need to review my exponent laws! $\endgroup$ – user2620463 Jan 14 '15 at 19:12
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First split the sum \begin{align}\sum\limits_{k=1}^n (\pi^k-3) &= \sum\limits_{k=1}^n\pi^k -\sum\limits_{k=1}^n 3\\ &= \sum\limits_{k=1}^n\pi^k -3\sum\limits_{k=1}^n 1\\ &= \sum\limits_{k=0}^{n-1}\pi^{k+1} -3\sum\limits_{k=1}^n 1\\ &= \pi\sum\limits_{k=0}^{n-1}\pi^{k} -3n,\end{align} where I shifted the index in the first sum. This first sum is a geometric series (with $a = 1$ and $r = \pi$). So we get $$\sum\limits_{k=1}^n (\pi^k-3) = \pi\frac{1-\pi^n}{1-\pi}-3n = \frac{\pi (\pi^n-1)}{\pi-1}-3n.$$

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$$\begin{align} \sum_{k=1}^n(\pi^k-3) & = \sum_{k=1}^n\pi^k-3\sum_{k-1}^n 1\\ & = \pi\sum_{k=1}^n\pi^{k-1}-3n\\ & = \pi\sum_{k=0}^{n-1}\pi^k-3n\\ & =\pi{\pi^n-1\over\pi-1}-3n. \end{align})$$

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\begin{align} \sum_{k=1}^n (\pi^k - 3) &= -3 n + \sum_{k=1}^n \pi^k = - 3n + (\pi + \pi^2 + \cdots + \pi^n) = \\ &= - 3n + \frac{(\pi - 1)(\pi + \pi^2 + \cdots + \pi^n)}{\pi - 1} =\\ &=- 3n + \frac{\pi^2 + \pi^3 + \cdots + \pi^n + \pi^{n+1} - \pi - \pi^2 - \pi^3 \cdots - \pi^n }{\pi - 1} =\\ &= -3n +\frac{\pi^{n+1} - \pi}{\pi - 1} = \frac{\pi(\pi^{n} - 1)}{\pi - 1}-3n \, . \end{align}

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