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Question Number 171620 by mokys last updated on 18/Jun/22

Commented by mokys last updated on 18/Jun/22

how can it proved

$${how}\:{can}\:{it}\:{proved} \\ $$

Commented by puissant last updated on 20/Jun/22

Σ_(i=1) ^n t^2 = t^2 Σ_(i=1) ^n = nt^2 .

$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{t}^{\mathrm{2}} =\:{t}^{\mathrm{2}} \:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:=\:{nt}^{\mathrm{2}} . \\ $$

Commented by kaivan.ahmadi last updated on 20/Jun/22

Σ_(i=1) ^n t^2 =1+2^2 +...+(n−1)^2 +n^2 = (Σ_(i=1) ^(n−1) t^2 )+n^2 =Σ_(i=1) ^(n−1) t^2 +n^2

$$\sum_{{i}=\mathrm{1}} ^{{n}} {t}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}^{\mathrm{2}} +...+\left({n}−\mathrm{1}\right)^{\mathrm{2}} +{n}^{\mathrm{2}} = \\ $$$$\left(\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} {t}^{\mathrm{2}} \right)+{n}^{\mathrm{2}} =\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} {t}^{\mathrm{2}} +{n}^{\mathrm{2}} \\ $$

Commented by puissant last updated on 20/Jun/22

Σ_(i=1) ^n i^2 = 1^2 +2^2 +....+(n−2)^2 +(n−1)^2 +n^2 = Σ_(i=1) ^(n−1) i^2 + n^2 = ((n(n+1)(2n+1))/6)

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

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