The sum of the infinite series frac{1}{9} + f | vedclass

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(A) The given series is $S = \frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + \dots \infty$. Multiplying by $2$,we get $2S =

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$\frac{1}{3}$

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(A) The given series is $S = \frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + \dots \infty$. Multiplying by $2$,we get $2S = \frac{2}{9} + \frac{2}{18} + \frac{2}{30} + \frac{2}{45} + \frac{2}{63} + \dots \infty$. This can be written as $2S = \frac{2}{3 	imes 3} + \frac{2}{3 	imes 6} + \frac{2}{6 	imes 5} + \dots$ which is not immediately helpful. Let us rewrite the terms as $T_n = \frac{1}{\frac{3n(n+2)}{2}} = \frac{2}{3n(n+2)}$. Using partial fractions,$T_n = \frac{2}{3} 	imes \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} ight) = \frac{1}{3} \left( \frac{1}{n} - \frac{1}{n+2} ight)$. For $n=1, 2, 3, \dots$,the terms are $\frac{1}{3} (1 - \frac{1}{3}), \frac{1}{3} (\frac{1}{2} - \frac{1}{4}), \frac{1}{3} (\frac{1}{3} - \frac{1}{5}), \dots$. Sum $S = \frac{1}{3} [ (1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{4} - \frac{1}{6}) + \dots ]$. Most terms cancel out,leaving $S = \frac{1}{3} (1 + \frac{1}{2}) = \frac{1}{3} (\frac{3}{2}) = \frac{1}{2}$. Wait,re-evaluating the series: $9, 18, 30, 45, 63$. The differences are $9, 12, 15, 18$. These are $3 	imes 3, 3 	imes 6, 3 	imes 10, 3 	imes 15, 3 	imes 21$. The denominators are $3 	imes \frac{n(n+1)}{2}$ where $n=2, 3, 4, 5, 6$. $T_n = \frac{2}{3n(n+1)}$. $S = \sum_{n=2}^{\infty} \frac{2}{3} (\frac{1}{n} - \frac{1}{n+1}) = \frac{2}{3} [(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots] = \frac{2}{3} 	imes \frac{1}{2} = \frac{1}{3}$. Thus,the correct option is $A$.

The sum of the infinite series $\frac{1}{9} + \frac{1}{18} + \frac{1}{30} + \frac{1}{45} + \frac{1}{63} + \dots \infty$ is equal to :-

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