NIMCET 2007 — Mathematics PYQ

Step 1: Recognize the Infinite Logarithmic Series

Let's factor out the common value $2$ from the given infinite series expansion:

$$S=2\left(\frac{1}{x}+\frac{1}{3x^3}+\frac{1}{5x^5}+\dots \right)$$

This exactly matches the standard logarithmic series expansion for odd power terms:

\frac{1}{2} \log \left( \frac{1 + t}{1 - t} \right) = t + \frac{t^3}{3} + \frac{t^5}{5} + \dots \quad \text{for } |t| < 1

By substituting $t=\frac{1}{x}$ into the standard formula, we get:

$$S=2\cdot \left[\frac{1}{2}\log \left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right)\right]$$

Simplify the expression inside the logarithm by taking a common denominator:

$$S=\log \left(\frac{\frac{x+1}{x}}{\frac{x-1}{x}}\right)=\log \left(\frac{x+1}{x-1}\right)$$

Step 2: Express $x$ in terms of $y$

We are given the algebraic relationship:

$$x(y^3-1)=1⟹x=\frac{1}{y^3-1}$$

Step 3: Substitute $x$ back into the Logarithmic Expression

Now, substitute $x=\frac{1}{y^3-1}$ into our simplified series sum equation $S=\log \left(\frac{x+1}{x-1}\right)$:

1. Evaluate the Numerator ($x+1$):

   $$x+1=\frac{1}{y^3-1}+1=\frac{1+(y^3-1)}{y^3-1}=\frac{y^3}{y^3-1}$$

2. Evaluate the Denominator ($x-1$):

   $$x-1=\frac{1}{y^3-1}-1=\frac{1-(y^3-1)}{y^3-1}=\frac{1-y^3+1}{y^3-1}=\frac{2-y^3}{y^3-1}$$

Step 4: Complete the Simplification

Combine the numerator and denominator fractions inside the logarithm:

$$S=\log \left[\frac{\frac{y^3}{y^3-1}}{\frac{2-y^3}{y^3-1}}\right]$$

Cancel out the common denominator term $(y^3-1)$:

$$S=\log \left[\frac{y^3}{2-y^3}\right]$$

Conclusion

The sum of the infinite series evaluates to $\log \left[\frac{y^3}{2-y^3}\right]$.

Correct Option: (a) $\log \left[\frac{y^3}{2-y^3}\right]$