The inverse-tangent function satisfies for

MATHEMATICS · Single correct · AP · Calculus BC · Infinite Sequences and Series · Infinite Sequences and Series

Problem

The inverse-tangent function satisfies arctan⁡x=∑n=0∞(−1)nx2n+12n+1\arctan x =\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{2n+1} for ∣x∣≤1|x|\le 1. Which of the following gives a power-series representation for 11+x2\dfrac{1}{1+x^{2}} valid for ∣x∣<1|x|<1? (a) ∑n=0∞(−1)nxn\displaystyle \sum_{n=0}^{\infty} (-1)^n x^{n} (b) ∑n=0∞(−1)n(2n+1)x2n\displaystyle \sum_{n=0}^{\infty} (-1)^n (2n+1) x^{2n} (c) ∑n=0∞(−1)nx2n2n+1\displaystyle \sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{2n+1} (d) ∑n=0∞(−1)nx2n\displaystyle \sum_{n=0}^{\infty} (-1)^n x^{2n}

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