題目 Problem 7.(10%) Use 11−x=∑n=0∞xn\dfrac{1}{1-x}=\sum_{n=0}^{\infty}x^n1−x1=∑n=0∞xn to express the value ∫01/212+x3 dx\int_{0}^{1/2}\frac{1}{2+x^3}\,dx∫01/22+x31dx as an infinite series. 解答 解法一 12+x3=∑n=0∞(−1)nx3n2n+1.\frac1{2+x^3}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{3n}}{2^{n+1}}.2+x31=n=0∑∞2n+1(−1)nx3n. 逐項積分到 1/21/21/2，得到 ∑n=0∞(−1)n24n+2(3n+1).\boxed{\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{4n+2}(3n+1)}}.n=0∑∞24n+2(3n+1)(−1)n.
