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Question Number 126617 by Khalmohmmad last updated on 22/Dec/20

1∙2+2∙3+3∙4+...+n(n+1)  S_n =?

$$\mathrm{1}\centerdot\mathrm{2}+\mathrm{2}\centerdot\mathrm{3}+\mathrm{3}\centerdot\mathrm{4}+...+{n}\left({n}+\mathrm{1}\right) \\ $$$${S}_{{n}} =? \\ $$

Answered by Dwaipayan Shikari last updated on 22/Dec/20

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

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

Commented by JDamian last updated on 22/Dec/20

Review the sum index

$${Review}\:{the}\:{sum}\:{index} \\ $$

Answered by Bird last updated on 23/Dec/20

S_n =Σ_(k=1) ^n k(k+1)  =Σ_(k=1) ^n  k^(2 )  +Σ_(k=1) ^n  k  =((n(n+1)(2n+1))/6) +((n(n+1))/2)  =((n(n+1))/2)(((2n+1)/3)+1)  =((n(n+1))/2)(((2n+4)/3))=((n(n+1)(n+2))/3)

$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} {k}\left({k}+\mathrm{1}\right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{2}\:} \:+\sum_{{k}=\mathrm{1}} ^{{n}} \:{k} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{3}}+\mathrm{1}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\left(\frac{\mathrm{2}{n}+\mathrm{4}}{\mathrm{3}}\right)=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{3}} \\ $$

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