AP Calculus BC 10.15 Representing Functions as Power Series - MCQs - Exam Style Questions
No-Calc Question
(B) \(\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{4n}}{(2n)!}\)
(C) \(\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{4n+1}}{(4n+1)(2n)!}\)
(D) \(\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}(4n)\,x^{4n-1}}{(2n)!}\)
▶️ Answer/Explanation
\(\cos u=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}u^{2n}}{(2n)!}\). With \(u=x^{2}\): \(\cos(x^{2})=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{4n}}{(2n)!}\).
Since \(f'(x)=\cos(x^{2})\) and \(f(0)=0\), \(f(x)=\displaystyle\int_{0}^{x}\cos(t^{2})\,dt=\int_{0}^{x}\sum_{n=0}^{\infty}\frac{(-1)^{n}t^{4n}}{(2n)!}\,dt\).
Integrate term-by-term: \(\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n)!}\cdot\frac{x^{4n+1}}{4n+1}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{4n+1}}{(4n+1)(2n)!}\).
✅ Answer: (C)
No-Calc Question
(B) The series converges, and \(\displaystyle \frac{2}{3}<\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}<1\).
(C) The series converges, and \(\displaystyle 1<\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}\).
(D) The series diverges.
▶️ Answer/Explanation
\(\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}=\arctan(1)=\frac{\pi}{4}\).
Since \(\displaystyle \frac{\pi}{4}\approx 0.7854\), we have \(\displaystyle \frac{2}{3}<\frac{\pi}{4}<1\).
Therefore the series converges and lies in the stated interval.
✅ Answer: (B)