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If $x$ is positive then the sum to infinity of the
series $\frac{1}{1+3 x}-\frac{1-3 x}{(1+3 x)^{2}}+\frac{(1-3 x)^{2}}{(1+3 x)^{3}}-\frac{(1-3 x)^{3}}{(1+3 x)^{4}}$ \ldots $\infty$ is
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series $\frac{1}{1+3 x}-\frac{1-3 x}{(1+3 x)^{2}}+\frac{(1-3 x)^{2}}{(1+3 x)^{3}}-\frac{(1-3 x)^{3}}{(1+3 x)^{4}}$ \ldots $\infty$ is
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The correct answer is:
$1 / 2$
The series is a G.P. with common ratio
$=\left(\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right)$ and $|\mathrm{r}|=\left|\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right|$ is less than 1 since $\mathrm{x}$ is
positive $\mathrm{S}_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\frac{1}{1+3 \mathrm{x}}}{1-\left\{-\left(\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right)\right\}}=\frac{1}{2}$
$=\left(\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right)$ and $|\mathrm{r}|=\left|\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right|$ is less than 1 since $\mathrm{x}$ is
positive $\mathrm{S}_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\frac{1}{1+3 \mathrm{x}}}{1-\left\{-\left(\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right)\right\}}=\frac{1}{2}$
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