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(A) Let $S = \frac{3}{2} x^{2} + \frac{5}{3} x^{3} + \frac{7}{4} x^{4} + \ldots \infty$We can rewrite the general term as $\frac{2n+1}{n} x^{n} = (2 +

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$x \left( \frac{1+x}{1-x} \right) - \log_{e}(1-x)$

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(A) Let $S = \frac{3}{2} x^{2} + \frac{5}{3} x^{3} + \frac{7}{4} x^{4} + \ldots \infty$We can rewrite the general term as $\frac{2n+1}{n} x^{n} = (2 + \frac{1}{n}) x^{n} = 2x^{n} + \frac{x^{n}}{n}$ for $n \geq 2$.Thus,$S = \sum_{n=2}^{\infty} (2x^{n} + \frac{x^{n}}{n}) = 2 \sum_{n=2}^{\infty} x^{n} + \sum_{n=2}^{\infty} \frac{x^{n}}{n}$.Using the geometric series sum $\sum_{n=2}^{\infty} x^{n} = \frac{x^{2}}{1-x}$ and the logarithmic expansion $-\log_{e}(1-x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \ldots$,we have $\sum_{n=2}^{\infty} \frac{x^{n}}{n} = -\log_{e}(1-x) - x$.Substituting these back: $S = 2 \left( \frac{x^{2}}{1-x} \right) - \log_{e}(1-x) - x$.$S = \frac{2x^{2} - x(1-x)}{1-x} - \log_{e}(1-x) = \frac{2x^{2} - x + x^{2}}{1-x} - \log_{e}(1-x) = \frac{3x^{2}-x}{1-x} - \log_{e}(1-x)$.Wait,re-evaluating the series: $S = \sum_{n=1}^{\infty} \frac{2n+1}{n+1} x^{n+1} = \sum_{n=1}^{\infty} (2 - \frac{1}{n+1}) x^{n+1} = 2 \sum_{n=1}^{\infty} x^{n+1} - \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$.$S = 2 \left( \frac{x^{2}}{1-x} \right) - (-\log_{e}(1-x) - x) = \frac{2x^{2}}{1-x} + x + \log_{e}(1-x) = \frac{2x^{2} + x - x^{2}}{1-x} + \log_{e}(1-x) = \frac{x^{2}+x}{1-x} + \log_{e}(1-x) = x \left( \frac{1+x}{1-x} \right) - \log_{e}(1-x)$.

If $0 < x < 1$,then $\frac{3}{2} x^{2} + \frac{5}{3} x^{3} + \frac{7}{4} x^{4} + \ldots$ is equal to:

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