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
$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\left(\overset{{n}} {\sum}{n}^{\mathrm{2}} +\overset{{n}} {\sum}\mathrm{2}{n}+\overset{{n}} {\sum}\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+{n}\left({n}+\mathrm{1}\right)+{n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\left({n}+\mathrm{1}\right)\left(\frac{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}}{\mathrm{6}}+{n}\right)+{n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{6}}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{1}\right)+{n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}.\mathrm{17}.\mathrm{609}+\mathrm{4} \\ $$$$ \\ $$
Commented by I want to learn more last updated on 06/Aug/20
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$