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Chapter 10第10章

Sequences & Series数列与级数

AP-Style Practice QuestionsAP 风格练习题

EASYMEDIUMHARD

Topics主题 10.1 - 10.15BC



Name:姓名:Period:课时:
PART ITopics 10.1 - 10.15专题 10.1 至 10.15

Multiple Choice Questions选择题

Show all supporting work on scratch paper. State clearly which convergence test you use and verify its conditions before applying it.请在草稿纸上写出所有解题步骤。使用某判别法前,须明确说明所用判别法并验证其条件。

Q1EASY10.1 Partial Sums (Telescoping)10.1 部分和(裂项相消)No Calculator

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}=$$\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}=$

Q2EASY10.2 Geometric Series10.2 等比级数No Calculator

$\displaystyle\sum_{n=1}^{\infty}\frac{5}{4^{n}}=$$\displaystyle\sum_{n=1}^{\infty}\frac{5}{4^{n}}=$

Q3EASY10.2 Geometric Series (Divergence)10.2 等比级数(发散)No Calculator

Which of the following series diverges?下列哪个级数发散?

Q4EASY10.3 nth Term Test10.3 第 n 项判别法No Calculator

$\displaystyle\sum_{n=1}^{\infty}\frac{2n^{2}+1}{3n^{2}+n}$$\displaystyle\sum_{n=1}^{\infty}\frac{2n^{2}+1}{3n^{2}+n}$

Q5MEDIUM10.4 Integral Test10.4 积分判别法No Calculator

$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$

Q6EASY10.5 $p$-Series10.5 $p$ 级数No Calculator

Which series converges?下列哪个级数收敛?

Q7MEDIUM10.5 Harmonic Series10.5 调和级数No Calculator

Which of the following is a true statement about the harmonic series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$?下列关于调和级数 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ 的说法中,哪个正确?

Q8EASY10.6 Direct Comparison Test10.6 比较判别法No Calculator

Since $0\le\dfrac{1}{n^{3}+1}\le\dfrac{1}{n^{3}}$ for all $n\ge 1$, and $\displaystyle\sum\frac{1}{n^{3}}$ converges, what does the Direct Comparison Test conclude about $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{3}+1}$?因对所有 $n\ge 1$ 有 $0\le\dfrac{1}{n^{3}+1}\le\dfrac{1}{n^{3}}$,且 $\displaystyle\sum\frac{1}{n^{3}}$ 收敛,比较判别法对 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{3}+1}$ 得出什么结论?

Q9MEDIUM10.6 Limit Comparison Test10.6 极限比较判别法No Calculator

$\displaystyle\sum_{n=1}^{\infty}\frac{n+2}{n^{3}-n+1}$$\displaystyle\sum_{n=1}^{\infty}\frac{n+2}{n^{3}-n+1}$

Q10MEDIUM10.7 Alternating Series Test10.7 交错级数判别法No Calculator

$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{n^{2}+1}$$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{n^{2}+1}$

Q11MEDIUM10.8 Ratio Test10.8 比值判别法No Calculator

$\displaystyle\sum_{n=1}^{\infty}\frac{2^{n}}{n!}$$\displaystyle\sum_{n=1}^{\infty}\frac{2^{n}}{n!}$

Q12MEDIUM10.9 Absolute vs. Conditional10.9 绝对收敛与条件收敛No Calculator

$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}$ is$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}$ 是

Q13HARD10.10 Alternating Series Error Bound10.10 交错级数误差界Calculator

For $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n!}$, what is the smallest $n$ such that the partial sum $S_n$ approximates the sum with error less than $0.001$?对 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n!}$,使部分和 $S_n$ 的误差小于 $0.001$ 的最小 $n$ 是多少?

Q14EASY10.11 Taylor Polynomials10.11 泰勒多项式No Calculator

In the 4th-degree Maclaurin polynomial for $f(x)=\cos x$, the coefficient of $x^{4}$ is在 $f(x)=\cos x$ 的 4 次麦克劳林多项式中,$x^{4}$ 的系数为

Q15HARD10.12 Lagrange Error Bound10.12 拉格朗日误差界Calculator

Let $P_3(x)$ be the 3rd-degree Taylor polynomial for $f(x)=e^{x}$ centered at $a=0$. If $|f^{(4)}(c)|\le 3$ for $0\le c\le 1$, the Lagrange error bound for $|f(1)-P_3(1)|$ is设 $P_3(x)$ 为 $f(x)=e^{x}$ 在 $a=0$ 处的 3 次泰勒多项式。若对 $0\le c\le 1$ 有 $|f^{(4)}(c)|\le 3$,则 $|f(1)-P_3(1)|$ 的拉格朗日误差界为

Q16MEDIUM10.13 Radius of Convergence10.13 收敛半径No Calculator

The radius of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x+1)^{n}}{n\cdot 3^{n}}$ is$\displaystyle\sum_{n=1}^{\infty}\frac{(x+1)^{n}}{n\cdot 3^{n}}$ 的收敛半径为

Q17MEDIUM10.14 Maclaurin Series10.14 麦克劳林级数No Calculator

Which series represents $\cos(x^{2})$?下列哪个级数表示 $\cos(x^{2})$?

Q18HARD10.15 Representing Functions as Series10.15 幂级数表示函数No Calculator

A power series representation of $\dfrac{x}{1-x^{3}}$ is$\dfrac{x}{1-x^{3}}$ 的幂级数表示为

PART IIShow All Work展示完整解题过程

Free-Response Questions自由回答题

Free-response answers must state the test used and verify its conditions explicitly. A calculator is permitted unless marked otherwise.自由回答题须明确说明所用判别法并验证其条件。除非特别注明,否则允许使用计算器。

FRQ 1EASY10.1 / 10.3 / 10.8 Classify Convergence10.1 / 10.3 / 10.8 判别收敛性No Calculator

For each series below, state whether it converges or diverges, name the test used, and (if it converges by a direct computation) give the sum.对下列各级数,判断其是收敛还是发散,写出所用判别法,并(若可直接求和)给出和。

(a) $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n+5}$
(b) $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+2)}$
(c) $\displaystyle\sum_{n=0}^{\infty}\frac{3^{n}}{n!}$
FRQ 2MEDIUM10.7 / 10.9 / 10.10 Alternating Series Analysis10.7 / 10.9 / 10.10 交错级数综合分析No Calculator

Let $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n+1}$.设 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n+1}$。

(a) Show that the series converges, verifying all three conditions of the Alternating Series Test.验证交错级数判别法的三个条件,证明该级数收敛。
(b) Determine whether the series is absolutely or conditionally convergent. Justify using an appropriate test on $\displaystyle\sum\left|a_n\right|$.判断该级数是绝对收敛还是条件收敛,并对 $\displaystyle\sum\left|a_n\right|$ 用适当的判别法加以论证。
(c) Use the alternating series error bound to find the smallest $n$ such that $S_n$ approximates the sum with error less than $0.05$.利用交错级数误差界,求使 $S_n$ 的误差小于 $0.05$ 的最小 $n$。
FRQ 3MEDIUM10.11 / 10.12 Taylor Polynomial & Error10.11 / 10.12 泰勒多项式与误差Calculator

Let $f(x)=\ln x$, and let $P_3(x)$ be the 3rd-degree Taylor polynomial for $f$ centered at $a=1$.设 $f(x)=\ln x$,$P_3(x)$ 为 $f$ 在 $a=1$ 处的 3 次泰勒多项式。

(a) Find $P_3(x)$. Show the derivatives used.求 $P_3(x)$,写出所用的各阶导数。
(b) Use $P_3(x)$ to approximate $\ln(1.2)$.用 $P_3(x)$ 近似 $\ln(1.2)$。
(c) Given that $|f^{(4)}(c)|\le 6$ for $1\le c\le 1.2$, find the Lagrange error bound for the approximation in part (b), and confirm it is consistent with the actual error ($\ln(1.2)\approx 0.18232$).已知对 $1\le c\le 1.2$ 有 $|f^{(4)}(c)|\le 6$,求 (b) 中近似值的拉格朗日误差界,并验证该界与实际误差一致($\ln(1.2)\approx 0.18232$)。
FRQ 4HARD10.13 Radius & Interval of Convergence10.13 收敛半径与收敛区间No Calculator

Let $\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^{n}}{n\cdot 5^{n}}$.设 $\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^{n}}{n\cdot 5^{n}}$。

(a) Use the ratio test to find the radius of convergence $R$.用比值判别法求收敛半径 $R$。
(b) State the open interval of convergence, then test each endpoint individually.写出开收敛区间,并分别检验各端点。
(c) State the interval of convergence.写出收敛区间。
(d) At the left endpoint, is the convergence absolute or conditional? Justify.在左端点处,级数是绝对收敛还是条件收敛?请说明理由。
FRQ 5HARD10.14 / 10.15 Maclaurin Series & Integration10.14 / 10.15 麦克劳林级数与积分Calculator

Let $g(x)=\ln(1+x)$.设 $g(x)=\ln(1+x)$。

(a) Starting from the geometric series for $\dfrac{1}{1+x}$, use term-by-term integration to find the Maclaurin series for $g(x)$.从 $\dfrac{1}{1+x}$ 的等比级数出发,用逐项积分求 $g(x)$ 的麦克劳林级数。
(b) Use the first four nonzero terms of this series to approximate $\ln(1.1)$.用该级数的前四个非零项近似 $\ln(1.1)$。
(c) Use the alternating series error bound to bound the error in the approximation in part (b).用交错级数误差界估计 (b) 中近似值的误差。
(d) State the radius of convergence of the series found in part (a), and determine whether $x=1$ is included in the interval of convergence.写出 (a) 中所求级数的收敛半径,并判断 $x=1$ 是否属于收敛区间。