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$1+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\ldots \ldots \infty$ is equal to
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The correct answer is:
$12$
It is an arithmetico-geometric series
$S_{\infty}=\frac{a}{1-r}+\frac{d r}{(1-r)^2}$ $=\frac{1}{1-\frac{1}{2}}+\frac{2}{\left(1-\frac{1}{2}\right)^2}=\frac{2}{\frac{1}{2}}+\frac{2}{\frac{1}{4}}$
$=4+8=12$.
$S_{\infty}=\frac{a}{1-r}+\frac{d r}{(1-r)^2}$ $=\frac{1}{1-\frac{1}{2}}+\frac{2}{\left(1-\frac{1}{2}\right)^2}=\frac{2}{\frac{1}{2}}+\frac{2}{\frac{1}{4}}$
$=4+8=12$.
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