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Question Number 157351 by physicstutes last updated on 22/Oct/21

$$\mathrm{Prove}\mathrm{by}\mathrm{mathematical}\mathrm{induction}$$$$\underset{r=1}{\sum ^n}\frac{1}{r(r+1)}=\frac{n}{n+1}$$

Commented by hknkrc46 last updated on 22/Oct/21

$$★\frac{1}{\mathit{r}(\mathit{r}+1)}=\frac{(\mathit{r}+1)-\mathit{r}}{\mathit{r}(\mathit{r}+1)}$$$$=\frac{\mathit{r}+1}{\mathit{r}(\mathit{r}+1)}-\frac{\mathit{r}}{\mathit{r}(\mathit{r}+1)}$$$$=\frac{1}{\mathit{r}}-\frac{1}{\mathit{r}+1}$$$$★\underset{r=1}{\sum ^n}\frac{1}{r(r+1)}=\underset{r=1}{\sum ^n}(\frac{1}{\mathit{r}}-\frac{1}{\mathit{r}+1})$$$$=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdot \cdot \cdot +(\frac{1}{\mathit{n}}-\frac{1}{\mathit{n}+1})$$$$=\frac{1}{1}-\frac{1}{\mathit{n}+1}=\frac{\mathit{n}}{\mathit{n}+1}$$

Answered by ajfour last updated on 22/Oct/21

$$S=\underset{r=1}{\sum ^n}\frac{1}{T_r}=\underset{r=1}{\sum ^n}(\frac{1}{r}-\frac{1}{r+1})$$$$=\underset{r=1}{\sum ^n}(t_r-t_{r+1})=1-\frac{1}{n+1}$$$$\Rightarrow S=\frac{n}{n+1}$$$$$$

Answered by physicstutes last updated on 22/Oct/21

$$\bullet \mathrm{prove}\mathrm{for}n=1.$$$$\mathrm{LHS}=\underset{r=1}{\sum ^1}\frac{1}{r(r+1)}=\frac{1}{1(1+1)}=\frac{1}{2}$$$$\mathrm{RHS}=\frac{1}{1+1}=\frac{1}{2}$$$$\Rightarrow \mathrm{true}\mathrm{for}n=1.$$$$\bullet \mathrm{Assume}\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}n=k$$$$\Rightarrow \underset{r=1}{\sum ^k}\frac{1}{r(r+1)}=\frac{k}{k+1}$$$$\bullet \mathrm{Prove}\mathrm{for}n=k+1.$$$$\underset{r=1}{\sum ^{k+1}}\frac{1}{r(r+1)}=\underset{r=1}{\sum ^k}\frac{1}{r(r+1)}+\frac{1}{(k+1)(k+2)}$$$$=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$$$$=\frac{1}{k+1}(k+\frac{1}{k+2})$$$$=\frac{1}{k+1}\left(\frac{k^2+2k+1}{k+2}\right)$$$$=\frac{1}{k+1}\left(\frac{{(k+1)}^2}{k+2}\right)$$$$=\frac{k+1}{k+2}$$$$\Rightarrow \mathrm{true}\mathrm{for}n=k+1$$$$\mathrm{hence}\mathrm{true}\mathrm{\forall }n\in \mathbb{Z}$$

Answered by som(math1967) last updated on 22/Oct/21

$$Toprove$$$$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$$$$p\left(1\right)L.H.S=\frac{1}{1\times 2}=\frac{1}{2}$$$$R.H.S=\frac{1}{1+1}=\frac{1}{2}$$$$\therefore trueforp\left(1\right)$$$$lettrueforp\left(m\right)$$$$\therefore \frac{1}{1\times 2}+\frac{1}{2\times 3}+...+\frac{1}{m(m+1)}=\frac{m}{m+1}$$$$nowp(m+1)$$$$=\frac{1}{1\times 2}+\frac{1}{2\times 3}+...+\frac{1}{m(m+1)}+\frac{1}{(m+1)(m+2)}$$$$=\frac{m}{m+1}+\frac{1}{(m+1)(m+2)}★$$$$=\frac{m(m+2)+1}{(m+1)(m+2)}$$$$=\frac{{(m+1)}^2}{(m+1)(m+2)}=\frac{m+1}{m+2}$$$$\therefore trueforp(m+1)$$$$\therefore \underset{\mathit{r}=1}{\sum ^{\mathit{n}}}\frac{1}{\mathit{r}(\mathit{r}+1)}=\frac{\mathit{n}}{\mathit{n}+1}$$$$★\frac{1}{1\times 2}+\frac{1}{2\times 3}+...+\frac{1}{m(m+1)}=\frac{m}{m+1}$$

Commented by peter frank last updated on 22/Oct/21

$$\mathrm{great}$$

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