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Question Number 96527 by student work last updated on 02/Jun/20

$$\mathrm{proof}\mathrm{that}{1}^2+2^2+3^2+....+{\mathrm{n}}^2=\frac{\mathrm{n}(2\mathrm{n}+1)(\mathrm{n}+1)}{6}$$

Commented by 06122004 last updated on 02/Jun/20

Answered by Farruxjano last updated on 02/Jun/20

Commented by student work last updated on 02/Jun/20

$$\mathrm{thanks}\mathrm{dear}\mathrm{great}$$

Answered by Sourav mridha last updated on 02/Jun/20

$$\mathit{t}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathit{a}\mathit{r}\mathit{e}\mathrm{many}\mathit{u}\mathit{s}\mathit{e}\mathit{f}\mathit{u}\mathit{l}\mathit{t}\mathit{e}\mathit{c}\mathit{h}\mathit{n}\mathit{i}\mathit{q}\mathit{u}\mathit{e}\mathrm{s}$$$$\mathit{b}\mathit{u}\mathit{t}\mathit{I}\mathit{w}\mathit{a}\mathit{n}\mathit{t}\mathit{t}\mathit{o}\mathit{s}\mathit{h}\mathit{o}\mathit{w}\mathit{v}\mathit{e}\mathit{r}\mathit{y}$$$$\mathit{u}\mathit{n}\mathit{u}\mathit{s}\mathit{u}\mathit{a}\mathit{l}\mathit{m}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{n}\mathit{d}...$$$$\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{E}\mathit{u}\mathit{l}\mathit{e}\mathit{r}-\mathit{M}\mathit{a}\mathit{c}\mathit{l}\mathit{a}\mathit{u}\mathit{r}\mathit{i}\mathit{n}\mathbf{formula}$$$$$$$$\underset{\mathit{m}=1}{\sum ^{\mathit{n}}}\mathit{f}\left(\mathit{m}\right)={\int }_1^{\mathit{n}}\mathit{f}\left(\mathit{x}\right)\mathit{d}\mathit{x}+\frac{1}{2}\mathit{f}\left(1\right)+\frac{1}{2}\mathbf{f}\left(\mathbf{n}\right)$$$$+\frac{{\mathbf{B}}_2}{2!}[{\mathit{f}}^{{}^{\prime }}\left(\mathit{n}\right)-{\mathit{f}}^{{}^{\prime }}\left(1\right)]+...........$$$$\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}{\mathit{B}}_{\mathit{n}}=\mathit{B}\mathit{e}\mathit{r}\mathit{n}\mathit{o}\mathit{u}\mathit{l}\mathit{l}\mathit{i}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}$$$$\mathit{a}\mathit{n}\mathit{d}\mathrm{here}{\mathrm{B}}_2=\frac{1}{6}...\mathit{f}\left(\mathit{m}\right)={\mathit{m}}^2...\mathit{s}\mathit{o}$$$$\mathrm{f}\left(1\right)=1\mathit{a}\mathit{n}\mathit{d}\mathit{f}\left(\mathit{n}\right)={\mathit{n}}^2$$$$\mathit{f}\left(\mathit{x}\right)={\mathit{x}}^2\mathit{s}\mathit{o}{\mathit{f}}^{{}^{\prime }}\left(\mathit{n}\right)=2\mathit{n},{\mathit{f}}^{{}^{\prime }}\left(1\right)=2$$$$$$$$\mathbf{so}\underset{\mathit{m}=1}{\sum ^{\mathit{n}}}{\mathit{m}}^2=\frac{{\mathit{n}}^3}{3}-\frac{1}{3}+\frac{1}{2}+\frac{{\mathit{n}}^2}{2}+\frac{(\mathit{n}-1)}{6}$$$$=\frac{2{\mathit{n}}^3+3{\mathit{n}}^2+\mathit{n}}{6}$$$$=\frac{\mathit{n}[2\mathit{n}(\mathit{n}+1)+1(\mathit{n}+1)]}{6}$$$$=\frac{\mathit{n}}{6}.(\mathit{n}+1)(2\mathit{n}+1).$$

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