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Question Number 65113 by rajesh4661kumar@gamil.com last updated on 25/Jul/19

Commented by Prithwish sen last updated on 25/Jul/19

$$\mathrm{No}.\mathrm{of}\mathrm{term}\mathrm{on}{\mathrm{n}}^{\mathrm{tn}}\mathrm{group}=2\mathrm{n}-1$$$$\mathrm{The}{1}^{\mathrm{st}}\mathrm{term}\mathrm{of}\mathrm{the}{\mathrm{n}}^{\mathrm{th}}\mathrm{group}$$$$=\frac{(\mathrm{n}-1)}{2}[2+(\mathrm{n}-2)2]+1={(\mathrm{n}-1)}^2+1$$$$={\mathrm{n}}^2-2\mathrm{n}+2$$$$\therefore \mathrm{the}\mathrm{sum}\mathrm{of}{\mathrm{n}}^{\mathrm{th}}\mathrm{group}$$$$=\frac{(2\mathrm{n}-1)}{2}[2({\mathrm{n}}^2-2\mathrm{n}+2)+(2\mathrm{n}-2)]$$$$=(2\mathrm{n}-1)({\mathrm{n}}^2-\mathrm{n}+1)={(\mathrm{n}-1)}^3+{\mathrm{n}}^3$$$$\mathrm{please}\mathrm{check}.$$

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