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I need to turn the following into something that doesn't use summation notation. Can someone help me figure out how to do that? I would know how to do it were it a simpler case but this one is difficult.

$$\sum_{i=0}^{\log n}5^{\log n-i}\left(3\cdot2^i+2\right)$$

My log's here are base 2.

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Please make sure that I correctly interpreted your expression. –  Brian M. Scott Oct 28 '12 at 22:54
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1 Answer

up vote 1 down vote accepted

$$\begin{align*} \sum_{i=0}^{\log n}5^{\log n-i}\left(3\cdot2^i+2\right)&=5^{\log n}\sum_{i=0}^{\log n}\left(3\left(\frac25\right)^i+2\left(\frac15\right)^i\right)\\ &=5^{\log n}\left(3\sum_{i=0}^{\log n}\left(\frac25\right)^i+2\sum_{i=0}^{\log n}\left(\frac15\right)^i\right)\;. \end{align*}$$

The last two summations are simple geometric series. Assuming that the upper limit of summation is actually $m=\lfloor\log n\rfloor$, they are

$$\sum_{i=0}^m\left(\frac25\right)^i=\frac{1-\left(\frac25\right)^{m+1}}{1-\frac25}=\frac53\left(1-\left(\frac25\right)^{m+1}\right)$$

and

$$\sum_{i=0}^m\left(\frac15\right)^i=\frac{1-\left(\frac15\right)^{m+1}}{1-\frac15}=\frac54\left(1-\left(\frac15\right)^{m+1}\right)\;.$$

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ohh ok, that works thanks. –  Jon Aird Oct 28 '12 at 23:57
    
@Jon: You’re welcome. –  Brian M. Scott Oct 29 '12 at 1:00
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