series summation

[xapzx]

How would I Evaluate this?

[MATH]\sum _{i=1}^m\:\left(\sum \:_{j=1}^n\left(i+j\right)\:\right)[/MATH]
Thank you

 

[MarkFL]

Hello, and welcome to FMH!

Let's look first at the inner sum:

[MATH]\sum_{j=1}^{n}(i+j)=i\sum_{j=1}^{n}(1)+\sum_{j=1}^{n}(j)=in+\frac{n(n+1)}{2}[/MATH]
And so now we have:

[MATH]\sum_{i=1}^{m}\left(in+\frac{n(n+1)}{2}\right)[/MATH]
Can you proceed?

 

[xapzx]

Ahh Yes, I see now.

Thank you very much!

 

[MarkFL]

To follow up:

[MATH]\sum_{i=1}^{m}\left(in+\frac{n(n+1)}{2}\right)=n\sum_{i=1}^{m}(i)+\frac{n(n+1)}{2}\sum_{i=1}^{m}(1)=\frac{nm(m+1)}{2}+\frac{nm(n+1)}{2}=\frac{mn}{2}\left(m+n+2\right)[/MATH]