A) 2
B) 3
C) 4
D) 6
Correct Answer: B
Solution :
Let\[S=1+\frac{2}{3}+\frac{6}{{{3}^{2}}}+\frac{10}{{{3}^{3}}}+\frac{14}{{{3}^{4}}}+....\infty \] \[\frac{\frac{S}{3}=\frac{1}{3}+\frac{2}{{{3}^{2}}}+\frac{6}{{{3}^{3}}}+\frac{10}{{{3}^{4}}}+....\infty }{\frac{2S}{3}=1+\frac{1}{3}+\frac{4}{{{3}^{2}}}+\frac{4}{{{3}^{3}}}+\frac{4}{{{3}^{4}}}+....\infty }\] \[=\frac{4}{3}+\frac{\frac{4}{{{3}^{2}}}}{1-\frac{1}{3}}\] \[=\frac{4}{3}+\frac{2}{3}=2\] \[\frac{2S}{3}=2\Rightarrow S=3\]
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