The sum of the infinite series (frac{1}{3}+fr | vedclass

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Question

(A) Let $a = \frac{4}{7}$ and $b = \frac{1}{3}$.Each term is of the form $\sum_{k=0}^{n} a^k b^{n-k} = \frac{a^{n+1} - b^{n+1}}{a - b}$.The series is

Vedclass · Accepted Answer

$ \frac{5}{2} $

Vedclass · Answer

(A) Let $a = \frac{4}{7}$ and $b = \frac{1}{3}$.Each term is of the form $\sum_{k=0}^{n} a^k b^{n-k} = \frac{a^{n+1} - b^{n+1}}{a - b}$.The series is $\sum_{n=1}^{\infty} \frac{a^{n+1} - b^{n+1}}{a - b} = \frac{1}{a - b} \left[ \sum_{n=1}^{\infty} a^{n+1} - \sum_{n=1}^{\infty} b^{n+1} ight]$.Here $a - b = \frac{4}{7} - \frac{1}{3} = \frac{12 - 7}{21} = \frac{5}{21}$.Sum $= \frac{21}{5} \left[ \frac{a^2}{1 - a} - \frac{b^2}{1 - b} ight] = \frac{21}{5} \left[ \frac{(4/7)^2}{1 - 4/7} - \frac{(1/3)^2}{1 - 1/3} ight]$.Sum $= \frac{21}{5} \left[ \frac{16/49}{3/7} - \frac{1/9}{2/3} ight] = \frac{21}{5} \left[ \frac{16}{21} - \frac{1}{6} ight]$.Sum $= \frac{21}{5} \left[ \frac{32 - 7}{42} ight] = \frac{21}{5} 	imes \frac{25}{42} = \frac{1}{1} 	imes \frac{5}{2} = \frac{5}{2}$.

The sum of the infinite series $(\frac{1}{3}+\frac{4}{7})+(\frac{1}{3^{2}}+\frac{1}{3}\times\frac{4}{7}+\frac{4^{2}}{7^{2}})+(\frac{1}{3^{3}}+\frac{1}{3^{2}}\times\frac{4}{7}+\frac{1}{3}\times\frac{4^{2}}{7^{2}}+\frac{4^{3}}{7^{3}}) + \dots$ is equal to -

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