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 have this question on my assignment for a computer science course (analysis of algorithms), so any help would be appreciated, but I am not looking for the answer itself.

I am trying to find the closed form solution for the following:


Since i = 5, n will always be $\ge5$, so I'm not sure why that's specified, unless I'm wrong about that.

The first thing I did was to factor out 3i:


Then I solved the inner sum:



$$\frac{3}{2}\sum_{i=5}^ni\,(i^2 + 2in + i + n^2 + n - 2)$$

I then tried to separate each term into different summations, combining terms treating n as a coefficient. I multiply each term by $12$ to get a common denominator, and then collect all the terms. The result I get is:


Which is almost correct.

Does anyone know why the coefficient would change from $\frac{3}{2}$ to $\frac{1}{8}$ while solving it? I don't understand that part.

share|cite|improve this question
up vote 4 down vote accepted

You are going to have terms such as $\sum_i i^2$ and $\sum_i i^3$. It turns out that there are simple expressions for these:

$$\sum_{i=1}^n i^2 = \frac{n (n+1)(2 n+1)}{6}$$

$$\sum_{i=1}^n i^3 = \left [ \frac{n (n+1)}{2} \right ]^2 $$

Collect terms in $i$, $i^2$, and $i^3$ in your sum and use these expressions. Shouldn't be too bad.

share|cite|improve this answer
Thanks! I forgot I could treat n as a coefficient. I need to remind myself that these nested summations are similar to partial derivatives in that respect. – MichaelThiessen Jan 25 '13 at 23:24

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.