Calculating Double Sum: (n=3)(i=0)∑(n=2)(j=0)∑(3i+2j)

[sapiental]

Homework Statement

Compute the following double sum

(n=3)(i=0)\sum(n=2)(j=0)\sum(3i+2j)

Homework Equations

sums

The Attempt at a Solution

my answer follows expanding the first sum, then just doing the last one

i get

(n=3)(i=0)\sum(6+9i) = 78<br /> <br /> thanks!

 

[sutupidmath]

\sum_{i=0}^{n=3} \sum_{j=0}^{n=2}(3i+2j)=\sum_{i=0}^{n=3}(6+9i)=78

Looks good to me.