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Question Number 23334 by math solver last updated on 28/Oct/17

Commented by math solver last updated on 28/Oct/17

$$\mathrm{find}\mathrm{sum}\mathrm{of}\mathrm{above}\mathrm{series}?$$

Commented by ajfour last updated on 28/Oct/17

$$infinitenumberoftermsor$$$$justnterms?$$

Commented by math solver last updated on 31/Oct/17

$$\mathrm{sir},\mathrm{actually}\mathrm{i}\mathrm{made}\mathrm{this}\mathrm{problem}$$$$\mathrm{myself}\mathrm{just}\mathrm{writing}{\mathrm{N}}_{\mathrm{r}}\mathrm{and}{\mathrm{D}}_{\mathrm{r}}$$$$\mathrm{in}\mathrm{A}.\mathrm{P}.\left(\mathrm{take}\mathrm{infinite}\mathrm{terms}\right)$$

Answered by ajfour last updated on 29/Oct/17

$$\mathit{S}=\mathit{n}-\frac{1}{\mathit{n}+1}-\frac{1}{\mathit{n}+2}+\frac{3}{2},$$$$because$$$$S=\underset{r=1}{\sum ^n}{T}_r=\underset{r=1}{\sum ^n}\frac{{(r+1)}^2+1}{{(r+1)}^2-1}$$$$=\frac{1}{2}\mathrm{\Sigma }\frac{{(r+2)}^2+r^2}{r(r+2)}$$$$=\frac{1}{2}\mathrm{\Sigma }(\frac{r+2}{r}+\frac{r}{r+2})$$$$=\frac{1}{2}\mathrm{\Sigma }(1+\frac{2}{r}+1-\frac{2}{r+2})$$$$=\underset{r=1}{\sum ^n}(1+\frac{1}{r}-\frac{1}{r+2})$$$$=n+(\frac{1}{1}-\frac{1}{3})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})$$$$+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})+...$$$$....+(\frac{1}{n-2}-\frac{1}{n})+(\frac{1}{n-1}-\frac{1}{n+1})$$$$+(\frac{1}{n}-\frac{1}{n+2})$$$$\Rightarrow S=n+\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}.$$

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