prove-that-1-r-1-n-r-1-2-n-n-1-2-r-1-n-r-2-1-6-n-n-1-2n-1-3-r-1-n-r-3-1-4-n-2-n-1-2-

Question Number 76368 by Rio Michael last updated on 26/Dec/19

prove that 1. Σ_(r=1) ^n r = (1/2)n(n+1) 2. Σ_(r=1) ^n r^2 = (1/6)n(n+1)(2n + 1) 3. Σ_(r=1) ^n r^3 = (1/4)n^2 (n + 1)^2

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

Commented by mr W last updated on 27/Dec/19

Commented by Rio Michael last updated on 27/Dec/19

$${thanks}\:{sir} \\ $$

Commented by john santu last updated on 27/Dec/19

$${hahahaha}\:\:\:……{complete}\:{sir} \\ $$

Commented by john santu last updated on 27/Dec/19

$${but}\:{sir}\:{the}\:{equation}\:{how}\:{to}\:{prove} \\ $$$${not}\:{list}\:{the}\:{formula}.\:{hahah}\:{great}\:{sir} \\ $$

Commented by mr W last updated on 27/Dec/19

there are many ways to calculate (to prove). if you already know S_k , then you can also get S_(k+1) .

$${there}\:{are}\:{many}\:{ways}\:{to}\:{calculate} \\ $$$$\left({to}\:{prove}\right).\:{if}\:{you}\:{already}\:{know}\:{S}_{{k}} , \\ $$$${then}\:{you}\:{can}\:{also}\:{get}\:{S}_{{k}+\mathrm{1}} . \\ $$

Answered by john santu last updated on 27/Dec/19

(1) (r+1)^2 −r^2 =2r+1 Σ_(r=1) ^n (r+1)^2 −r^2 =Σ_(r=1) ^n (2r+1) (n+1)^2 −1=2Σ_(r=1) ^n (r)+n n^2 +n=2Σ_(r=1) ^n r Σ_(r=1) ^n r=((n^2 +n)/2) ■

$$\left(\mathrm{1}\right)\:\left({r}+\mathrm{1}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{2}{r}+\mathrm{1} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left({r}+\mathrm{1}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}{r}+\mathrm{1}\right) \\ $$$$\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{2}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left({r}\right)+{n} \\ $$$${n}^{\mathrm{2}} +{n}=\mathrm{2}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}=\frac{{n}^{\mathrm{2}} +{n}}{\mathrm{2}}\:\blacksquare \\ $$$$ \\ $$

Answered by john santu last updated on 27/Dec/19

(2)(r+1)^3 −r^3 =3r^2 +3r+1 Σ_(r=1) ^n {(r+1)^3 −r^3 }=Σ_(r=1) ^n (3r^2 +3r+1) (n+1)^3 −n^3 =3Σ_(r=1) ^n r^2 + ((3n(n+1))/2)+n Σ_(r=1) ^n r^2 =((n(n+1)(2n+1))/6) ■

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

Answered by benjo last updated on 27/Dec/19

prove (3) use (r+1)^4 −r^(4 ) = 4r^3 +6r^2 +4r+1

$$\mathrm{prove}\:\left(\mathrm{3}\right)\:\mathrm{use}\:\left(\mathrm{r}+\mathrm{1}\right)^{\mathrm{4}} \:−\mathrm{r}^{\mathrm{4}\:} \:=\:\mathrm{4r}^{\mathrm{3}} +\mathrm{6r}^{\mathrm{2}} \:+\mathrm{4r}+\mathrm{1} \\ $$

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