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The sum to infinity of the series $S = 1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \frac{14}{3^{4}} + \dots$ is

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Correct option is D

Let S=1+23+632+1033+1434+… 13S=13+232+633+1034+…Subtracting, we get 23S=1+13+432+433+434+… =1+13+4/321−1/3=43+23=2⇒ S=3

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The sum to infinity of the series S=1+23+632+1033+1434+…is

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The sum to infinity of the series $S = 1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \frac{14}{3^{4}} + \dots$ is