A) \[\frac{9}{2}\]
B) \[\frac{27}{8}\]
C) \[\frac{81}{8}\]
D) \[\frac{81}{4}\]
Correct Answer: B
Solution :
\[\frac{{{T}_{r+1}}}{{{T}_{r}}}=\frac{1}{3}\] If \[{{T}_{1}}=a;\,\,{{T}_{2}}=\frac{a}{3},\,\,{{T}_{3}}=\frac{a}{{{3}^{2}}},\,\,.....\] \[\therefore \]\[{{T}_{7}}=\frac{a}{{{3}^{6}}}=\frac{1}{243}\Rightarrow \frac{a}{{{3}^{5}}\times 3}=\frac{1}{{{3}^{5}}}\Rightarrow a=3\] \[\therefore \]\[{{T}_{r}}.{{T}_{r+1}}=\frac{a}{{{3}^{r-1}}}\times \frac{a}{{{3}^{r}}}=\frac{{{a}^{2}}}{{{3}^{2r-1}}}=\frac{{{3}^{2}}}{{{3}^{2r-1}}}\] \[\Rightarrow \]\[\sum\limits_{r=1}^{\infty }{{{T}_{r}}.{{T}_{r+1}}=\left( 3+\frac{1}{3}+\frac{1}{{{3}^{3}}}+\frac{1}{{{3}^{5}}}+..........\infty \right)}\] \[=\frac{3}{1-\frac{2}{{{3}^{2}}}}=\frac{27}{8}\]
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