Question Number 65113 by rajesh4661kumar@gamil.com last updated on 25/Jul/19
Commented by Prithwish sen last updated on 25/Jul/19
![No. of term on n^(tn) group = 2n−1 The 1^(st ) term of the n^(th ) group = (((n−1))/2)[2+(n−2)2] +1 = (n−1)^2 +1 =n^2 −2n+2 ∴ the sum of n^(th) group =(((2n−1))/2)[2(n^2 −2n+2)+(2n−2)] = (2n−1)(n^2 −n+1) = (n−1)^3 + n^3 please check.](https://www.tinkutara.com/question/Q65127.png)
$$\mathrm{No}.\:\mathrm{of}\:\mathrm{term}\:\mathrm{on}\:\mathrm{n}^{\mathrm{tn}} \mathrm{group}\:=\:\mathrm{2n}−\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{1}^{\mathrm{st}\:} \mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{n}^{\mathrm{th}\:} \mathrm{group} \\ $$$$=\:\frac{\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}}\left[\mathrm{2}+\left(\mathrm{n}−\mathrm{2}\right)\mathrm{2}\right]\:+\mathrm{1}\:=\:\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$=\mathrm{n}^{\mathrm{2}} −\mathrm{2n}+\mathrm{2} \\ $$$$\therefore\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{n}^{\mathrm{th}} \:\mathrm{group} \\ $$$$=\frac{\left(\mathrm{2n}−\mathrm{1}\right)}{\mathrm{2}}\left[\mathrm{2}\left(\mathrm{n}^{\mathrm{2}} −\mathrm{2n}+\mathrm{2}\right)+\left(\mathrm{2n}−\mathrm{2}\right)\right] \\ $$$$=\:\left(\mathrm{2n}−\mathrm{1}\right)\left(\mathrm{n}^{\mathrm{2}} −\mathrm{n}+\mathrm{1}\right)\:=\:\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{3}} \:+\:\mathrm{n}^{\mathrm{3}} \\ $$$$\mathrm{please}\:\mathrm{check}. \\ $$