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Question Number 151037 by qaz last updated on 17/Aug/21

$$\underset{\mathrm{r}=1,\mathrm{r}\ne \mathrm{s}}{\sum ^{\mathrm{n}}}\underset{\mathrm{s}=1}{\sum ^{\mathrm{n}}}\frac{\mathrm{rs}}{\mathrm{n}(\mathrm{n}-1)}\stackrel{?}{=}\frac{(\mathrm{n}+1)(3\mathrm{n}+2)}{12}$$

Answered by mindispower last updated on 17/Aug/21

$$=\underset{r=1}{\sum ^{r=n}}\underset{s=1}{\sum ^n}\frac{rs}{n(n-1)}-\underset{s=1}{\sum ^n}\frac{s^2}{n(n-1)}$$$$=\frac{n^2{(n+1)}^2}{4n(n-1)}-\frac{n(2n+1)(n+1)}{6n(n-1)}$$$$=\frac{n(n+1)(3n(n+1)-2(2n+1))}{12n(n-1)}$$$$=\frac{(n+1)(3n^2-n-2)}{12(n-1)}=\frac{n+1}{12(n-1)}.3(n-1)(n+\frac{2}{3})$$$$=\frac{(n+1)(3n+2)}{12}$$

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