1. ## Evaluating summations

I'm still struggling with basic summation evaluation.

I've managed to get a formula like this:

$\displaystyle \sum_{k=2}^n \sum_{j=1}^{3n} 1$

But the 3n is confusing me and I'm not sure what my next step should be.

I think I evaluate down to :

$\displaystyle \sum_{k=2}^n(3n-j)$? and then where?

2. Don't know what the summand exactly is.

What's $\displaystyle n1$ ?

3. Oops sorry, that was just a missing bracket in latex. Should be fixed now.

4. That makes sense, but the inner sum is actually $\displaystyle 3n.$

5. Thanks for the help, but I'm still a little fuzzy. I'm not sure what you mean by the inner sum is actually 3n?

6. You're respected to know that is $\displaystyle \sum_{j=1}^n1=n,$ if the upper index of the sum were $\displaystyle 3n,$ what's the new value of the sum?

7. Originally Posted by Krizalid
You're respected to know that is $\displaystyle \sum_{j=1}^n1=n,$ if the upper index of the sum were $\displaystyle 3n,$ what's the new value of the sum?

Ok, so it should evaluate next to

$\displaystyle \sum_{k=2}^n3n$

then

3$\displaystyle \sum_{k=2}^nn^2$ ?

Sorry my summation skills are really fuzzy.

8. That sum is w.r.t. $\displaystyle k,$ then you can pull back that $\displaystyle 3n$ and you'll have $\displaystyle \sum_{k=2}^n1=\sum_{k=1}^{n-1}1=n-1.$

9. Originally Posted by Krizalid
then you can pull back that $\displaystyle 3n$ and you'll have $\displaystyle \sum_{k=2}^n1=\sum_{k=1}^{n-1}1=n-1.$

Sorry again, but I'm not sure what you mean by pull back that 3n?