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- Homework Statement:
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I computed the double sum

$$\sum_{i=0}^n\sum_{j=i+1}^n j = \sum_{i=0}^n\big(\frac{n(n+1)}{2}-\frac{i(i+1)}{2}\big)=\frac{n(n+1)(2n+1)}{6}$$

and realized the double sum is equal to $$\sum_{i=1}^ni^2$$

which leads to

$$\sum_{i=0}^n\sum_{j=i+1}^n j = \sum_{i=1}^ni^2$$

Is there a proof of this equality that doesn't require computing out the sums?

I thought that I should try changing the boundaries of integration

##0\leq i+1\leq j\leq n## so

$$\sum_{i=0}^n\sum_{j=i+1}^n j = \sum_{j=0}^n\sum_{i=0}^j j = \sum_{j=0}^n j(j+1)=\Big(\frac{n(n+1)}{2}+1\Big)\Big(\frac{n(n+1)}{2}\Big)$$

but I was unable to even do this properly since the above isn't equal to ##\frac{n(n+1)(2n+1)}{6}##

$$\sum_{i=0}^n\sum_{j=i+1}^n j = \sum_{i=0}^n\big(\frac{n(n+1)}{2}-\frac{i(i+1)}{2}\big)=\frac{n(n+1)(2n+1)}{6}$$

and realized the double sum is equal to $$\sum_{i=1}^ni^2$$

which leads to

$$\sum_{i=0}^n\sum_{j=i+1}^n j = \sum_{i=1}^ni^2$$

Is there a proof of this equality that doesn't require computing out the sums?

I thought that I should try changing the boundaries of integration

##0\leq i+1\leq j\leq n## so

$$\sum_{i=0}^n\sum_{j=i+1}^n j = \sum_{j=0}^n\sum_{i=0}^j j = \sum_{j=0}^n j(j+1)=\Big(\frac{n(n+1)}{2}+1\Big)\Big(\frac{n(n+1)}{2}\Big)$$

but I was unable to even do this properly since the above isn't equal to ##\frac{n(n+1)(2n+1)}{6}##

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