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This is the first time I encountered a double summation. $$\sum_{i=1}^j\sum_{j=1}^5 3ij$$ I have solved the right summation but I don't know what to do with the 'j' upper limit of the left summation. Perhaps its order need to be changed first?

Martin Sleziak
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JamesLp
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  • Yes, left summation needs to be done first. You can solve this and answer yourself. – cosmo5 Nov 20 '20 at 08:17
  • I would say that the notation is incorrect. The upper limit $j$ of the first sum is supposed to be bound to (and hence dependent on) the second sum, which should not happen. – Sangchul Lee Nov 26 '20 at 20:28

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$$S=\sum_{i=1}^j\sum_{j=1}^5 3ij=\sum_{j=1}^5\sum_{i=1}^j 3ij= \sum_{j=1}^{5}3j \sum_{i=1}^{j} i=\sum_{j=1}^{5}[3j(j(j+1)/2].=\frac{1}{2}\sum_{j=1}^{5}[3j^3+3j^2] $$ $$\implies S=\frac{3}{2} \left(\frac{5(5+1)}{2}\right)^2+\frac{3}{2}\frac{5(5+1)(2.5+1)}{6}=420$$

Z Ahmed
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