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Q: Wait, why did we split the coefficient into $2 - \frac{1}{n}$? How was I supposed to see that?

Great question! In JEE problems, whenever you see a sequence of fractions like $3/2, 5/3, 7/4$, always check the difference between the numerator and denominator or look for a constant offset. Here, $3/2 = 1.5$, $5/3 \approx 1.66$, $7/4 = 1.75$. They are approaching 2. By writing $2 - 1/n$, we decompose a complex term into two simple, recognizable series.

Q: Why can't I just use the formula for the sum of an infinite series directly?

Because this isn't a simple geometric progression! It is a 'mixed' series. The coefficients change with $n$, which breaks the standard $a/(1-r)$ rule. We must decompose it into parts that *are* standard series.

Q: I keep getting the sign wrong in the logarithmic part. Why is it $-\ln(1-x) - x$?

The expansion is $-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$. If your series starts at $\frac{x^2}{2}$, you have to isolate that part by moving the $x$ to the other side. So, $\frac{x^2}{2} + \frac{x^3}{3} + \dots = -\ln(1-x) - x$. Always write out the first few terms of the standard expansion to compare!

Q: Is there a constraint on $x$ that I should worry about?

Yes, the problem states $0 < x < 1$. This is crucial because the geometric series $x^2 + x^3 + \dots$ only converges if the common ratio $|x| < 1$. Without this constraint, the sum would be undefined.

Q: How do I know when to stop simplifying the final expression?

Always look at your options! The options are in a specific factored form. Once you reach $\frac{x^2+x}{1-x} + \ln(1-x)$, look at the first option: $x(\frac{1+x}{1-x}) + \ln(1-x)$. It matches perfectly, so that is your cue to stop.

Question

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

Options:

(A) $x \left( \frac{1+x}{1-x} \right) + \log_e(1 - x)$

(B) $x \left( \frac{1-x}{1+x} \right) + \log_e(1 - x)$

(C) $\frac{1-x}{1+x} + \log_e(1 - x)$

(D) $\frac{1+x}{1-x} + \log_e(1 - x)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The infinite series is given by $S = \frac{3}{2}x^2 + \frac{5}{3}x^3 + \frac{7}{4}x^4 + \dots$. At first glance, this sequence does not appear to be a simple geometric progression.

However, in the context of JEE Advanced, complexity is often just simplicity in disguise. We must look for the underlying pattern in the coefficients.

### The Insight of Decomposition

Observe the coefficients: $\frac{3}{2}, \frac{5}{3}, \frac{7}{4}, \dots$. The general term for the $n$-th term (starting from $n=2$) is $\frac{2n-1}{n}x^n$.

We can perform an algebraic split on the coefficient:
$$\frac{2n-1}{n} = 2 - \frac{1}{n}$$

This allows us to transform the entire series $S$ into two distinct, manageable entities:
$$S = \sum_{n=2}^{\infty} \left( 2 - \frac{1}{n} \right) x^n = 2 \sum_{n=2}^{\infty} x^n - \sum_{n=2}^{\infty} \frac{x^n}{n}$$

### Taming the Geometric Beast

Let us focus on the first part, $S_1 = 2(x^2 + x^3 + x^4 + \dots)$. This is a classic infinite geometric progression with first term $a = x^2$ and common ratio $r = x$.

Given the constraint $0 < x < 1$, we apply the sum formula $S_{\infty} = \frac{a}{1-r}$:
$$S_1 = 2 \left( \frac{x^2}{1-x} \right) = \frac{2x^2}{1-x}$$

This expression represents the backbone of our series, accounting for the constant '2' in our decomposition.

### The Logarithmic Secret

Now, consider the second part, $S_2 = -(\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots)$. This series is related to the Taylor expansion of the natural logarithm:
$$-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$$

Our series $S_2$ is missing the first term, $x$. Therefore, we adjust the expansion:
$$\frac{x^2}{2} + \frac{x^3}{3} + \dots = -\ln(1-x) - x$$

Applying the negative sign from our original split, we obtain:
$$S_2 = -(-\ln(1-x) - x) = \ln(1-x) + x$$

### The Grand Synthesis

We now combine our two components, $S_1$ and $S_2$, to find the total sum:
$$S = \frac{2x^2}{1-x} + x + \ln(1-x)$$

To simplify the algebraic portion, we combine the terms over a common denominator:
$$\frac{2x^2 + x(1-x)}{1-x} = \frac{2x^2 + x - x^2}{1-x} = \frac{x^2 + x}{1-x}$$

Factoring out $x$, we arrive at the final, elegant form of the series:
**$$S = x \left( \frac{1+x}{1-x} \right) + \ln(1-x)$$**

Animated Solution for Mathematics - Sequence and Series: If $0 < x < 1$ , then $\frac{3}{2} x^{2} + \frac{5}{3} x^{3} + \frac{7}{4} x^{4} + \dots$ , is equal to:

Select Answer:

Visualized Solution

Identifying the Infinite Series $S$

Analyzing the Coefficients $\frac{2 n - 1}{n}$

Splitting into Two Series

The Geometric Series Part

Summing the Infinite G.P.

The Logarithmic Series Part

Adjusting the Standard Expansion

Summing the Logarithmic Part

Combining the Results

Algebraic Simplification

The Final Result

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Insight of Decomposition

Taming the Geometric Beast

The Logarithmic Secret

The Grand Synthesis

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 0<x<1, then 23​x2+35​x3+47​x4+…, is equal to: