A) \[\frac{1}{2}\left[ 1+\frac{1}{1.3.5......(2n+1)} \right]\]
B) \[\frac{1}{2}\left[ 1-\frac{1}{2.4.6......2n} \right]\]
C) \[\frac{1}{2}\left[ 1-\frac{1}{1.3.5......(2n+1)} \right]\]
D) None of these
Correct Answer: C
Solution :
| \[{{t}_{n}}=\frac{n}{1.3.5......(2n+1)}\]\[=\frac{1}{2}\frac{(2n+1)-1}{1.3.5....(2n+1)}\] | |
| \[=\frac{1}{2}\left[ \frac{1}{1.3.5....(2n+1)}-\frac{1}{1.3.5.....(2n+1)} \right]\] | |
| \[=\frac{1}{2}({{T}_{n-1}}-{{T}_{n}})\] | |
| \[\therefore \,\,\,2{{t}_{n}}={{T}_{n-1}}-{{T}_{n}}\] | ?? (i) |
| \[\left[ where\,{{T}_{n}}=\frac{1}{1.3.5....(2n+1)} \right]\] |
| \[2{{S}_{n}}=\sum\limits_{n=2}^{n}{2{{t}_{n}}}+2{{t}_{1}}\] |
| \[=({{T}_{1}}-{{T}_{2}})+({{T}_{2}}-{{T}_{3}})+...+{{T}_{n-1}}-{{T}_{n}}+2{{t}_{1}}\]\[\Rightarrow 2\,({{S}_{n}}-{{t}_{1}})={{T}_{1}}-{{T}_{n}}\]\[\Rightarrow 2{{S}_{n}}=2{{t}_{1}}+{{T}_{1}}-{{T}_{n}}\]\[\Rightarrow 2{{S}_{n}}=2.\,\,\frac{1}{1.3}+\frac{1}{1.3}-\frac{1}{1.3.5....(2n+1)}\] \[\Rightarrow {{S}_{n}}=\frac{1}{2}\left[ \frac{1}{1.3.5....(2n+1)} \right]\] |
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