r-1-r-s-n-s-1-n-rs-n-n-1-n-1-3n-2-12-

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

\underset{r = 1, r \neq s}{\sum^{n}} \underset{s = 1}{\sum^{n}} \frac{rs}{n (n - 1)} \overset{?}{=} \frac{(n + 1) (3 n + 2)}{12}

Answered by mindispower last updated on 17/Aug/21

= \underset{r = 1}{\sum^{r = n}} \underset{s = 1}{\sum^{n}} \frac{r s}{n (n - 1)} - \underset{s = 1}{\sum^{n}} \frac{s^{2}}{n (n - 1)}

= \frac{n^{2} {(n + 1)}^{2}}{4 n (n - 1)} - \frac{n (2 n + 1) (n + 1)}{6 n (n - 1)}

= \frac{n (n + 1) (3 n (n + 1) - 2 (2 n + 1))}{12 n (n - 1)}

= \frac{(n + 1) (3 n^{2} - n - 2)}{12 (n - 1)} = \frac{n + 1}{12 (n - 1)} . 3 (n - 1) (n + \frac{2}{3})

= \frac{(n + 1) (3 n + 2)}{12}