If the r^{th} term of a series is (2r + 1)2^{ | vedclass

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(C) The $r^{th}$ term is given by $T_r = (2r + 1)2^{-r} = \frac{2r + 1}{2^r}$.The sum of infinite terms is $S_{\infty} = \sum_{r=1}^{\infty} \frac{2r

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$5$

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(C) The $r^{th}$ term is given by $T_r = (2r + 1)2^{-r} = \frac{2r + 1}{2^r}$.The sum of infinite terms is $S_{\infty} = \sum_{r=1}^{\infty} \frac{2r + 1}{2^r} = \sum_{r=1}^{\infty} \frac{2r}{2^r} + \sum_{r=1}^{\infty} \frac{1}{2^r}$.Let $S = \sum_{r=1}^{\infty} \frac{2r + 1}{2^r} = \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \frac{9}{16} + \dots$This is an arithmetico-geometric series. Multiplying by $\frac{1}{2}$:$\frac{1}{2}S = \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + \dots$Subtracting the two equations:$S - \frac{1}{2}S = \frac{3}{2} + (\frac{5}{4} - \frac{3}{4}) + (\frac{7}{8} - \frac{5}{8}) + \dots$$\frac{1}{2}S = \frac{3}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + \dots$$\frac{1}{2}S = \frac{3}{2} + 2(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots)$The term in the bracket is a geometric series with $a = \frac{1}{4}$ and $r = \frac{1}{2}$,so the sum is $\frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.$\frac{1}{2}S = \frac{3}{2} + 2(\frac{1}{2}) = \frac{3}{2} + 1 = \frac{5}{2}$.Therefore,$S = 5$.

If the $r^{th}$ term of a series is $(2r + 1)2^{-r}$,what is the sum of its infinite terms?

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