A) 2
B) 3
C) 4
D) 6
Correct Answer: B
Solution :
Let S be the sum of the given series. i.e.\[S=1+\frac{2}{3}+\frac{6}{{{3}^{2}}}+\frac{10}{{{3}^{3}}}+...\] \[\Rightarrow \]\[(S-1)=\frac{2}{3}+\frac{6}{{{3}^{2}}}+\frac{6}{{{3}^{2}}}+\frac{10}{{{3}^{3}}}+...\] ?(i) \[\Rightarrow \]\[(S-1)\times \frac{1}{3}=\frac{2}{{{3}^{2}}}+\frac{6}{{{3}^{3}}}+\frac{10}{{{3}^{4}}}+...\] On subtracting Eq. (ii) from Eq. (i), we get \[\frac{1}{3}(S-1)=\frac{2}{3}+\frac{4}{{{3}^{2}}}+\frac{4}{{{3}^{3}}}+\frac{4}{{{3}^{4}}}+...\] \[\Rightarrow \]\[\frac{2}{3}(S-1)=\frac{2}{3}+\frac{\frac{4}{{{3}^{2}}}}{1-\frac{1}{3}}\] \[\Rightarrow \]\[\frac{2}{3}(S-1)=\frac{2}{3}+\frac{1}{3}\]\[\Rightarrow \]\[S-1=2\]\[\Rightarrow \]\[S=3\]
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