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Question Number 15671 by tawa tawa last updated on 12/Jun/17

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

Answered by icyfalcon999 last updated on 12/Jun/17

$$1)\mathrm{proving}\mathrm{that}\mathrm{the}\mathrm{statement}\mathrm{true}\mathrm{when}\mathrm{n}=1$$$$\mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{2\left(1\right)}{1+1}=\frac{2}{2}=1=\mathrm{L}.\mathrm{H}.\mathrm{S}.$$$$2)\mathrm{suppose}\mathrm{that}\mathrm{the}\mathrm{statement}\mathrm{is}\mathrm{true}\mathrm{when}\mathrm{n}=\mathrm{k},\mathrm{k}\in \mathbb{N}\mathrm{u}$$$$$$$$$$$$1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...\mathrm{k}}=\frac{2\mathrm{k}}{\mathrm{k}+1}$$$$3)\mathrm{proving}\mathrm{that}\mathrm{the}\mathrm{statement}\mathrm{true}\mathrm{when}\mathrm{n}=\mathrm{k}+1$$$$1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...\mathrm{k}}+\frac{1}{1+2+3+...+\mathrm{k}+1}=\frac{2(\mathrm{k}+1)}{\mathrm{k}+2}$$$$\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{1}{1+2+3+...+\mathrm{k}+1}$$$$=\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{1}{\frac{(\mathrm{k}+1)(\mathrm{k}+2)}{2}}$$$$=\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{2}{(\mathrm{k}+1)(\mathrm{k}+2)}$$$$=\frac{2\mathrm{k}(\mathrm{k}+2)+2}{(\mathrm{k}+1)(\mathrm{k}+2)}$$$$=\frac{{2\mathrm{k}}^2+4\mathrm{k}+2}{(\mathrm{k}+1)(\mathrm{k}+2)}$$$$=\frac{2({\mathrm{k}}^2+2\mathrm{k}+1)}{(\mathrm{k}+1)(\mathrm{k}+2)}$$$$=\frac{2{(\mathrm{k}+1)}^2}{(\mathrm{k}+1)(\mathrm{k}+2)}$$$$=\frac{2(\mathrm{k}+1)}{(\mathrm{k}+2)}$$$$=\mathrm{R}.\mathrm{H}.\mathrm{S}.$$$$\mathrm{from}1,2,3\mathrm{the}\mathrm{statment}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}\mathrm{natural}\mathrm{numbers}$$

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