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

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SUMMARY

The double sum calculation for the expression (n=3)(i=0)∑(n=2)(j=0)∑(3i+2j) results in a total of 78. The solution involves expanding the first sum and simplifying the second sum, leading to the final result of 78. The calculations confirm that the approach taken is correct, as verified by the participant's detailed breakdown of the sums.

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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]
 
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[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.
 

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