Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Please make sure that I correctly interpreted your expression. – Brian M. Scott Oct 28 '12 at 22:54
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




share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.