answer: 1/24\n#____ - use series to approxima...

# Explanation: ## Step1: Recall cosine - series expansion The Maclaurin series for $\cos t=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}t^{2n}=1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\frac{t^{6}}{6!}+\cdots$. Let $t = x^{2}$, then $\cos(x^{2})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}(x^{2})^{2n}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{4n}=1-\frac{x^{4}}{2!}+\frac{x^{8}}{4!}-\frac{x^{12}}{6!}+\cdots$. ## Step2: Integrate the series term - by - term $\int_{0}^{1}\cos(x^{2})dx=\int_{0}^{1}\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{4n}dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}\int_{0}^{1}x^{4n}dx$. Since $\int_{0}^{1}x^{4n}dx=\left[\frac{x^{4n + 1}}{4n+1}\right]_{0}^{1}=\frac{1}{4n + 1}$, we have $\int_{0}^{1}\cos(x^{2})dx=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!(4n + 1)}=1-\frac{1}{2!(5)}+\frac{1}{4!(9)}-\frac{1}{6!(13)}+\cdots$. ## Step3: Determine the number of terms for error bound The series $\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!(4n + 1)}$ is an alternating series. For an alternating series $\sum_{n = 0}^{\infty}(-1)^{n}a_{n}$ ($a_{n}\gt0$, $a_{n+1}\leq a_{n}$, $\lim_{n\rightarrow\infty}a_{n}=0$), the error bound $|R_N|\leq a_{N + 1}$. We want $|R_N|\lt\frac{1}{100}$. Let's calculate the terms: When $n = 0$, $a_{0}=\frac{1}{(0)!(4\times0 + 1)} = 1$. When $n = 1$, $a_{1}=\frac{1}{2!(4\times1+1)}=\frac{1}{2\times5}=\frac{1}{10}$. When $n = 2$, $a_{2}=\frac{1}{4!(4\times2 + 1)}=\frac{1}{24\times9}=\frac{1}{216}\lt\frac{1}{100}$. ## Step4: Calculate the approximation $\int_{0}^{1}\cos(x^{2})dx\approx1-\frac{1}{10}+\frac{1}{216}=\frac{216 - 21.6+1}{216}=\frac{195.4}{216}\approx0.905$. The approximation using the first three terms of the series (i.e., $n = 0,1,2$) is $\sum_{n = 0}^{2}\frac{(-1)^{n}}{(2n)!(4n + 1)}=1-\frac{1}{10}+\frac{1}{216}=\frac{216-21.6 + 1}{216}=\frac{195.4}{216}\approx0.905$. # Answer: $\sum_{n = 0}^{2}\frac{(-1)^{n}}{(2n)!(4n + 1)}=1-\frac{1}{10}+\frac{1}{216}$