1 + frac{3}{2} + frac{5}{2^2} + frac{7}{2^3} | vedclass

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(B) The given series is an arithmetico-geometric series: $S = 1 + \frac{3}{2} + \frac{5}{2^2} + \frac{7}{2^3} + \dots \infty$.Here,the arithmetic part

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The given series is an arithmetico-geometric series: $S = 1 + \frac{3}{2} + \frac{5}{2^2} + \frac{7}{2^3} + \dots \infty$.Here,the arithmetic part is $1, 3, 5, 7, \dots$ with first term $a = 1$ and common difference $d = 2$.The geometric part is $1, \frac{1}{2}, \frac{1}{2^2}, \frac{1}{2^3}, \dots$ with common ratio $r = \frac{1}{2}$.The sum of an infinite arithmetico-geometric series is given by $S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$.Substituting the values: $S_{\infty} = \frac{1}{1 - \frac{1}{2}} + \frac{2 	imes \frac{1}{2}}{(1 - \frac{1}{2})^2}$.$S_{\infty} = \frac{1}{\frac{1}{2}} + \frac{1}{(\frac{1}{2})^2} = 2 + \frac{1}{\frac{1}{4}} = 2 + 4 = 6$.Wait,let us re-evaluate: $S = 1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \dots$$\frac{1}{2}S = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \dots$Subtracting: $S - \frac{1}{2}S = 1 + (\frac{3}{2} - \frac{1}{2}) + (\frac{5}{4} - \frac{3}{4}) + (\frac{7}{8} - \frac{5}{8}) + \dots$$\frac{1}{2}S = 1 + 1 + \frac{2}{4} + \frac{2}{8} + \dots = 1 + (1 + \frac{1}{2} + \frac{1}{4} + \dots) = 1 + \frac{1}{1 - \frac{1}{2}} = 1 + 2 = 3$.Therefore,$S = 3 	imes 2 = 6$.

$1 + \frac{3}{2} + \frac{5}{2^2} + \frac{7}{2^3} + \dots \infty$ is equal to

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