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Question Number 63428 by Rio Michael last updated on 04/Jul/19
It is given that S_n =Σ_(r=1) ^n (3r^(2 ) −3r−1). Use the the formulae of Σ_(r=1) ^n r^(2 ) and Σ_(r=1) ^n r to show that S_n =n^3 . sir Forkum Michael
$${It}\:{is}\:{given}\:{that}\:{S}_{{n}} =\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}{r}^{\mathrm{2}\:} −\mathrm{3}{r}−\mathrm{1}\right).\:{Use}\:{the}\:{the}\:{formulae} \\ $$$${of}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{2}\:\:} {and}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}\:\:{to}\:{show}\:{that}\:{S}_{{n}} ={n}^{\mathrm{3}} . \\ $$$${sir}\:{Forkum}\:{Michael} \\ $$
Commented by Tony Lin last updated on 04/Jul/19
Σ_(r=1) ^n (3r^2 −3r−1) =3Σ_(r=1) ^n r^2 −3Σ_(r=1) ^n r−Σ_(r=1) ^n 1 =((3n(n+1)(2n+1))/6)−((3n(n+1))/2)−n =((n(n+1)(2n+1)−3n(n+1)−2n)/2) =(((2n^3 +3n^2 +n)−(3n^2 +3n)−2n)/2) ⇒the question may be Σ_(r=1) ^n (3r^2 −3r+1) then(((2n^3 +3n^2 +n)−(3n^2 +3n)+2n)/2) =n^3
$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}{r}^{\mathrm{2}} −\mathrm{3}{r}−\mathrm{1}\right) \\ $$$$=\mathrm{3}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{2}} −\mathrm{3}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}−\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1} \\ $$$$=\frac{\mathrm{3}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−\frac{\mathrm{3}{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−{n} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{3}{n}\left({n}+\mathrm{1}\right)−\mathrm{2}{n}}{\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{2}{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{n}\right)−\left(\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}\right)−\mathrm{2}{n}}{\mathrm{2}} \\ $$$$\Rightarrow{the}\:{question}\:{may}\:{be}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{3}{r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{1}\right) \\ $$$${then}\frac{\left(\mathrm{2}{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +{n}\right)−\left(\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}\right)+\mathrm{2}{n}}{\mathrm{2}} \\ $$$$={n}^{\mathrm{3}} \\ $$
Commented by Rio Michael last updated on 04/Jul/19
correct
$${correct} \\ $$

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