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

S=1+23+632+1033+1434+…∞13S=13+232+633+1034+…∞23S=1+13+432+433+434+…∞23S=43+43.13+43.132+43.133+…∞above series infinite G.P then23S=431−13 ⇒23S=2 ⇒S=3

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

S=1+23+632+1033+1434+…∞13S=13+232+633+1034+…∞23S=1+13+432+433+434+…∞23S=43+43.13+43.132+43.133+…∞above series infinite G.P then23S=431−13 ⇒23S=2 ⇒S=3

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

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$3$

$2$

$1$

$4$

answer is C.

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Detailed Solution

$\begin{array}{l} S = 1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \frac{14}{3^{4}} + \dots \infty \\ \frac{1}{3} S = \frac{1}{3} + \frac{2}{3^{2}} + \frac{6}{3^{3}} + \frac{10}{3^{4}} + \dots \infty \\ \frac{2}{3} S = (1 + \frac{1}{3}) + \frac{4}{3^{2}} + \frac{4}{3^{3}} + \frac{4}{3^{4}} + \dots \infty \\ \frac{2}{3} S = \frac{4}{3} + \frac{4}{3} . \frac{1}{3} + \frac{4}{3} . \frac{1}{3^{2}} + \frac{4}{3} . \frac{1}{3^{3}} + \dots \infty \\ a b o v e s e r i e s i n f i n i t e G . P t h e n \\ \frac{2}{3} S = \frac{\frac{4}{3}}{1 - \frac{1}{3}} \Rightarrow \frac{2}{3} S = 2 \Rightarrow S = 3 \end{array}$