Sum the series to infinity frac{3}{4} - frac{ | vedclass

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Question

(C) The given series is $S = \left( \frac{3}{4} + \frac{3}{4^3} + \frac{3}{4^5} + \dots \right) - \left( \frac{5}{4^2} + \frac{5}{4^4} + \frac{5}{4^6}

Vedclass · Accepted Answer

$\frac{7}{15}$

Vedclass · Answer

(C) The given series is $S = \left( \frac{3}{4} + \frac{3}{4^3} + \frac{3}{4^5} + \dots ight) - \left( \frac{5}{4^2} + \frac{5}{4^4} + \frac{5}{4^6} + \dots ight)$.This consists of two infinite geometric series.For the first series,the first term $a_1 = \frac{3}{4}$ and the common ratio $r_1 = \frac{1}{4^2} = \frac{1}{16}$.The sum $S_1 = \frac{a_1}{1 - r_1} = \frac{3/4}{1 - 1/16} = \frac{3/4}{15/16} = \frac{3}{4} 	imes \frac{16}{15} = \frac{4}{5}$.For the second series,the first term $a_2 = \frac{5}{4^2} = \frac{5}{16}$ and the common ratio $r_2 = \frac{1}{4^2} = \frac{1}{16}$.The sum $S_2 = \frac{a_2}{1 - r_2} = \frac{5/16}{1 - 1/16} = \frac{5/16}{15/16} = \frac{5}{16} 	imes \frac{16}{15} = \frac{1}{3}$.Therefore,the total sum $S = S_1 - S_2 = \frac{4}{5} - \frac{1}{3} = \frac{12 - 5}{15} = \frac{7}{15}$.

Sum the series to infinity $\frac{3}{4} - \frac{5}{4^2} + \frac{3}{4^3} - \frac{5}{4^4} + \frac{3}{4^5} - \frac{5}{4^6} + \dots$

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