A) 2
B) 3
C) 4
D) 6
Correct Answer: B
Solution :
[b]: Let \[S=1+\frac{2}{3}+\frac{6}{{{3}^{2}}}+\frac{10}{{{3}^{3}}}+\frac{14}{{{3}^{4}}}+....\infty \] \[\Rightarrow \]\[S-1=\frac{2}{3}+\frac{6}{{{3}^{2}}}+\frac{10}{{{3}^{3}}}+\frac{14}{{{3}^{4}}}+....\infty \] ?(i) \[\Rightarrow \]\[(S-1)\times \frac{1}{3}=\frac{2}{{{3}^{2}}}+\frac{6}{{{3}^{3}}}+\frac{10}{{{3}^{4}}}+\frac{14}{{{3}^{5}}}+....\infty \]?(ii) Subtracting (ii) from (i), we get \[\frac{2}{3}(S-1)=\frac{2}{3}+\frac{4}{{{3}^{2}}}+\frac{4}{{{3}^{3}}}+\frac{4}{{{3}^{4}}}+....\infty \] \[\Rightarrow \]\[\frac{2}{3}(S-1)=\frac{2}{3}+\frac{\frac{4}{{{3}^{2}}}}{1-\frac{1}{3}}\] \[\Rightarrow \]\[\frac{2}{3}(S-1)=\frac{2}{3}+\frac{2}{3}\Rightarrow S=3\].
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