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Question Number 11854 by tawa last updated on 02/Apr/17

$$\mathrm{Solve}\mathrm{by}\mathrm{mathematical}\mathrm{induction}\mathrm{that}$$$$1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+\mathrm{n}}=\frac{2\mathrm{n}}{\mathrm{n}+1}$$

Answered by sandy_suhendra last updated on 03/Apr/17

$$\mathrm{for}\mathrm{n}=1$$$$1=\frac{2.1}{1+1}\left(\mathrm{is}\mathrm{true}\right)$$$$$$$$\mathrm{for}\mathrm{n}=\mathrm{k}$$$$1+\frac{1}{1+2}+...+\frac{1}{1+2+3+...+\mathrm{k}}=\frac{2\mathrm{k}}{\mathrm{k}+1}$$$$$$$$\mathrm{for}\mathrm{n}=(\mathrm{k}+1)\mathrm{should}\mathrm{be}=\frac{2(\mathrm{k}+1)}{[(\mathrm{k}+1)+1]}$$$$[1+\frac{1}{1+2}+...+\frac{1}{1+2+3...+\mathrm{k}}]+\frac{1}{1+2+3+...+\mathrm{k}+(\mathrm{k}+1)}$$$$=\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{1}{\frac{\mathrm{k}+1}{2}(1+\mathrm{k}+1)}$$$$=\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{2}{(\mathrm{k}+1)(\mathrm{k}+2)}$$$$=\frac{2\mathrm{k}(\mathrm{k}+2)+2}{(\mathrm{k}+1)(\mathrm{k}+2)}$$$$=\frac{{2\mathrm{k}}^2+4\mathrm{k}+2}{(\mathrm{k}+1)(\mathrm{k}+2)}$$$$=\frac{2(\mathrm{k}+1)(\mathrm{k}+1)}{(\mathrm{k}+1)[(\mathrm{k}+1)+1]}$$$$=\frac{2(\mathrm{k}+1)}{[(\mathrm{k}+1)+1]}\left(\mathrm{is}\mathrm{proved}\right)$$

Commented by Mr Chheang Chantria last updated on 03/Apr/17

$$\mathit{p}\mathit{e}\mathit{r}\mathit{f}\mathit{e}\mathit{c}\mathit{t}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}.$$

Commented by tawa last updated on 03/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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