How can I find the result of:
$\sum\limits_{i=1}^{n}\prod\limits_{j=1}^{2} ij$
I know that $\prod\limits_{j=1}^{2} ij = 2i^2$, so I should simply do the summation as $\sum\limits_{i=1}^n2i^2$?
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$\begingroup$ $$\frac{1}{3} n (n+1) (2 n+1)$$ $\endgroup$– David G. StorkCommented Jan 12, 2021 at 18:40
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$\begingroup$ yes, if you copied the problem correctly that seems pretty clear $\sum\limits_{i}^n (\prod\limits_{j=1}^2 ij) =\sum\limits_{i}^n (1j\cdot 2j)= \sum\limits_{i}^n (2j^2) = 2\sum_{i=1}^n i^2$ and you should be able know what that is. ... I guess I'm wondering how this rather strange expression came about in the first place. $\endgroup$– fleabloodCommented Jan 12, 2021 at 19:17
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$\begingroup$ One could generalize that $\sum\limits_{i=1}^n \prod\limits{j=1}^k ij =k! \sum\limits_{i=1}^n i^k$.... I guess..... But writing something of the form $\prod\limits_j ij$ seems fairly weird to me. as that'd just be $\prod\limits_{j=1}^k ij = i^k \prod j = i^k k!$. $\endgroup$– fleabloodCommented Jan 12, 2021 at 19:22
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1 Answer
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This should be easy:
$$ \sum_{i=1}^n 2i^2 = 2 \sum_{i=1}^n i^2 = 2 \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{3}$$