Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 106730 by I want to learn more last updated on 06/Aug/20

Answered by Dwaipayan Shikari last updated on 06/Aug/20

Σ_(n=1) ^n ((n^2 (n+1)^2 )/(4n^2 ))=(1/4)(Σ^n n^2 +Σ^n 2n+Σ^n 1) =(1/4)(((n(n+1)(2n+1))/6)+n(n+1)+n) =(1/4)((n+1)(((2n^2 +1)/6)+n)+n) =(1/4)((1/6)(n+1)(2n^2 +6n+1)+n) =(1/(24)).17.609+4

$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\left(\overset{{n}} {\sum}{n}^{\mathrm{2}} +\overset{{n}} {\sum}\mathrm{2}{n}+\overset{{n}} {\sum}\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+{n}\left({n}+\mathrm{1}\right)+{n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\left({n}+\mathrm{1}\right)\left(\frac{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}}{\mathrm{6}}+{n}\right)+{n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{6}}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{1}\right)+{n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}.\mathrm{17}.\mathrm{609}+\mathrm{4} \\ $$$$ \\ $$

Commented by I want to learn more last updated on 06/Aug/20

Thanks sir.

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com