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

Σ_(r=1,r≠s) ^n Σ_(s=1) ^n ((rs)/(n(n−1)))=^? (((n+1)(3n+2))/(12))

$$\underset{\mathrm{r}=\mathrm{1},\mathrm{r}\neq\mathrm{s}} {\overset{\mathrm{n}} {\sum}}\:\:\underset{\mathrm{s}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{rs}}{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}\overset{?} {=}\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{3n}+\mathrm{2}\right)}{\mathrm{12}} \\ $$

Answered by mindispower last updated on 17/Aug/21

=Σ_(r=1) ^(r=n) Σ_(s=1) ^n ((rs)/(n(n−1)))−Σ_(s=1) ^n (s^2 /(n(n−1))) =((n^2 (n+1)^2 )/(4n(n−1)))−((n(2n+1)(n+1))/(6n(n−1))) =((n(n+1)(3n(n+1)−2(2n+1)))/(12n(n−1))) =(((n+1)(3n^2 −n−2))/(12(n−1))) =((n+1)/(12(n−1))).3(n−1)(n+(2/3)) =(((n+1)(3n+2))/(12))

$$=\underset{{r}=\mathrm{1}} {\overset{{r}={n}} {\sum}}\underset{{s}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{rs}}{{n}\left({n}−\mathrm{1}\right)}−\underset{{s}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{s}^{\mathrm{2}} }{{n}\left({n}−\mathrm{1}\right)} \\ $$$$=\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}{n}\left({n}−\mathrm{1}\right)}−\frac{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)}{\mathrm{6}{n}\left({n}−\mathrm{1}\right)} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{3}{n}\left({n}+\mathrm{1}\right)−\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)\right)}{\mathrm{12}{n}\left({n}−\mathrm{1}\right)} \\ $$$$=\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{3}{n}^{\mathrm{2}} −{n}−\mathrm{2}\right)}{\mathrm{12}\left({n}−\mathrm{1}\right)}\:=\frac{{n}+\mathrm{1}}{\mathrm{12}\left({n}−\mathrm{1}\right)}.\mathrm{3}\left({n}−\mathrm{1}\right)\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$=\frac{\left({n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}{\mathrm{12}} \\ $$

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