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

  • Thread starter Thread starter sapiental
  • Start date Start date
  • Tags Tags
    Sum
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 8K views
sapiental
Messages
110
Reaction score
0

Homework Statement



Compute the following double sum

(n=3)(i=0)[tex]\sum[/tex](n=2)(j=0)[tex]\sum[/tex](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)[tex]\sum[tex](6+9i) = 78<br /> <br /> thanks![/tex][/tex]
 
Physics news on Phys.org
[tex]\sum_{i=0}^{n=3} \sum_{j=0}^{n=2}(3i+2j)=\sum_{i=0}^{n=3}(6+9i)=78[/tex]

Looks good to me.