Key Concepts and Formulas

This problem involves the summation of an infinite series. We will utilize the following key concepts:

1. Maclaurin Series for $\ln(1-x)$: The Maclaurin series expansion for $\ln(1-x)$ is given by: $\ln(1-x) = - \left( x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots \right) \quad \text{for } |x| < 1$ Rearranging this, we get: $-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots \quad \text{for } |x| < 1$
2. Sum of an Infinite Geometric Series: The sum of an infinite geometric series $a + ar + ar^2 + \dots$ with first term $a$ and common ratio $r$ is: $S_\infty = \frac{a}{1-r} \quad \text{for } |r| < 1$ The condition $0 < x < 1$ ensures that all series involved are convergent.

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Step-by-Step Solution

Let the given series be $S$. $S = {3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + \dots \dots \infty$

Step 1: Identify the General Term of the Series We examine the structure of the terms in the series. The first term is $\frac{3}{2}x^2$. The second term is $\frac{5}{3}x^3$. The third term is $\frac{7}{4}x^4$. If we consider the $n$-th term corresponding to $x^n$ (where $n$ starts from 2), the numerator is $2n-1$ and the denominator is $n$. Thus, the general term can be expressed as $T_n = \frac{2n-1}{n}x^n$ for $n \ge 2$.

Step 2: Rewrite the General Term to Facilitate Splitting We can simplify the coefficient $\frac{2n-1}{n}$: $\frac{2n-1}{n} = \frac{2n}{n} - \frac{1}{n} = 2 - \frac{1}{n}$ Substituting this back into the general term, we get $T_n = \left(2 - \frac{1}{n}\right)x^n$.

Step 3: Split the Series into Two Separate Series Using the rewritten general term, we can expand the series $S$: $S = \sum_{n=2}^{\infty} \left(2 - \frac{1}{n}\right)x^n$ This sum can be split into two separate sums: $S = \sum_{n=2}^{\infty} 2x^n - \sum_{n=2}^{\infty} \frac{1}{n}x^n$ $S = 2 \sum_{n=2}^{\infty} x^n - \sum_{n=2}^{\infty} \frac{x^n}{n}$

Step 4: Evaluate the First Series (Geometric Series) The first part is $2 \sum_{n=2}^{\infty} x^n = 2(x^2 + x^3 + x^4 + \dots)$. This is an infinite geometric series with the first term $a = x^2$ and the common ratio $r = x$. Since $0 < x < 1$, $|r| < 1$, and the sum converges to: $x^2 + x^3 + x^4 + \dots = \frac{a}{1-r} = \frac{x^2}{1-x}$ So, the first part of $S$ is: $2 \left( \frac{x^2}{1-x} \right) = \frac{2x^2}{1-x}$

Step 5: Evaluate the Second Series (Related to Logarithmic Series) The second part is $-\sum_{n=2}^{\infty} \frac{x^n}{n} = -\left(\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots \right)$. We know the Maclaurin series for $-\ln(1-x)$: $-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots$ The series we have, $\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots$, is the Maclaurin series for $-\ln(1-x)$ with the first term $x$ removed. Therefore, $\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots = -\ln(1-x) - x$ The second part of $S$ is: $-\left(-\ln(1-x) - x\right) = \ln(1-x) + x$

Step 6: Combine the Results and Simplify Now, we add the results from Step 4 and Step 5 to find $S$: $S = \frac{2x^2}{1-x} + (\ln(1-x) + x)$ $S = \frac{2x^2}{1-x} + x + \ln(1-x)$ To combine the terms with $x$, we find a common denominator: $S = \frac{2x^2}{1-x} + \frac{x(1-x)}{1-x} + \ln(1-x)$ $S = \frac{2x^2 + x - x^2}{1-x} + \ln(1-x)$ $S = \frac{x^2 + x}{1-x} + \ln(1-x)$ Factor out $x$ from the numerator: $S = \frac{x(x+1)}{1-x} + \ln(1-x)$ This can be written as: $S = x \left( \frac{1+x}{1-x} \right) + \ln(1-x)$

This expression matches option (A).

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Common Mistakes & Tips

• Index of Series: Be careful with the starting index of the series. The given series starts with $x^2$, while the standard logarithmic series starts with $x$. This requires adjustment by adding or subtracting terms.
• Algebraic Manipulation: Ensure accuracy when combining fractions or simplifying expressions. A small error in algebra can lead to an incorrect final answer.
• Convergence Conditions: Always verify that the condition on $x$ (here, $0 < x < 1$) satisfies the convergence criteria for the series being used.

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Summary

The problem requires summing an infinite series. We first identified the general term of the series and rewrote it by splitting the coefficient into a constant and a reciprocal term. This allowed us to decompose the given series into a geometric series and a series related to the logarithmic expansion. By evaluating each part separately and then combining them, we arrived at the final simplified expression.

The final answer is $\boxed{x\left( {{{1 + x} \over {1 - x}}} \right) + {\log _e}(1 - x)}$, which corresponds to option (A).