4
$\begingroup$

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:

$$\sum_{i=5}^n\sum_{j=2}^{n+i}3ji,\;for\;n\ge5$$

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:

$$\sum_{i=5}^n3i\sum_{j=2}^{n+i}j$$

Then I solved the inner sum:

$$\sum_{i=5}^n3i\;\left(\,\sum_{j=1}^{n+1}j-\sum_{j=1}^1j\,\right)$$

$$\sum_{i=5}^n3i\left(\frac{(n+i)(n+i+1)}{2}-1\right)$$

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

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:

$$\frac{3}{2}\left(17n^4+34n^3-113n^2-850n-1320\right)$$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – MichaelThiessen Jan 25 '13 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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