# Calculate !

• Oct 9th 2013, 11:30 AM
Kazey
Calculate !
Good afternoon everyone !

I'm supposed to calculate this sum !

Attachment 29427

I know that in the end, i'm supposed to get something like : (n(n+1)(4n-1))/6 or if you dev. this expression : 5n/6...

Could anybody give me the step-by-step resolution of this exercice ? I have really no idea about what i'm supposed to do with it. I tried to transform it into a double sum, but that's all...

• Oct 9th 2013, 12:18 PM
topsquark
Re: Calculate !
Quote:

Originally Posted by Kazey
Good afternoon everyone !

I'm supposed to calculate this sum !

Attachment 29427

I know that in the end, i'm supposed to get something like : (n(n+1)(4n-1))/6 or if you dev. this expression : 5n/6...

Could anybody give me the step-by-step resolution of this exercice ? I have really no idea about what i'm supposed to do with it. I tried to transform it into a double sum, but that's all...

You aren't sure if n(n + 1)(4n - 1)/6 is right?? Well, it's correct.

I'd do it inductively. Your base case is n = 1 and it works just fine.

So assume that $\sum_{1 \leq i, j \leq k} max(i, j) = \frac{k(k + 1)(4k - 1)}{6}$ for some value of k.

Then you need to show that
$\sum_{1 \leq i, j \leq k + 1} max(i, j) = \frac{(k + 1)((k + 1) + 1)(4(k + 1) - 1)}{6}$

So. I'll give you an outline, let's see what you can do with it. You need to show that
$\sum_{1 \leq i, j \leq k + 1} max(i, j) = \sum_{1 \leq i, j \leq k} max(i, j) + \text{Other terms}$

How many terms do you need to add to the k case on the RHS to make it equal to the k + 1 case?

-Dan