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Question Number 45364 by Tawa1 last updated on 12/Oct/18

$$\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{n}\mathrm{terms}:\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{(2\mathrm{n}-1)(2\mathrm{n}+1)}=?$$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

$$T_n=\frac{1}{(2n+1)(2n-1)}$$$$T_n=\frac{1}{2}\left[\frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)}\right]$$$$T_n=\frac{1}{2}[\frac{1}{2n-1}-\frac{1}{2n+1}]$$$$T_1=\frac{1}{2}[\frac{1}{1}-\frac{1}{3}]$$$$T_2=\frac{1}{2}[\frac{1}{3}-\frac{1}{5}]$$$$T_3=\frac{1}{2}[\frac{1}{5}-\frac{1}{7}]$$$$....$$$$...$$$$addthem$$$$S_n=\frac{1}{2}[1-\frac{1}{2n+1}]$$

Commented by Tawa1 last updated on 12/Oct/18

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

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