Question

For \( -1<x<1 \), the sum of the power series \[ 1 + \sum_{n=2}^{\infty} (-1)^{n-1} n^2 x^{n-1} \text{ is} \]

• A. \( \frac{1 - x}{(1 + x^3)} \)
• B. \( \frac{1 + x^2}{(1 + x^4)} \)
• C. \( \frac{1 - x}{(1 + x^2)} \)
• D. \( \frac{(1 + x^2)^3}{(1 + x^3)} \)

Hint:

When dealing with power series, look for standard expansions or known series forms to simplify the summation.

Answer 1

The Correct Option is A

Step 1: Recognize the power series form.
The series \( 1 + \sum_{n=2}^{\infty} (-1)^{n-1} n^2 x^{n-1} \) is a modified power series of the form: \[ S(x) = \sum_{n=1}^{\infty} (-1)^{n-1} a_n x^{n}. \]
Step 2: Identify the closed-form of the series.
We recognize that the sum of this series is a standard series expansion, and through analysis and simplification, the closed-form expression of the sum turns out to be: \[ S(x) = \frac{1 - x}{1 + x^2}. \]
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{(C)} \).