BC Calculus Series Test

BC Calculus Series Test

12th Grade

15 Qs

quiz-placeholder

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BC Calculus Series Test

BC Calculus Series Test

Assessment

Quiz

Other

12th Grade

Medium

Created by

Jennifer Denn

Used 18+ times

FREE Resource

15 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

The coefficient of the

 x3x^3  term in the Taylor series for  e2xe^{2x}  centered at x = 0 is

1/6

1/3

2/3

4/3

2.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

 n=1(13)2n\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^{2n}  =



1/8

1/3

1

9/8

 \infty  

3.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

The coefficient of  (x1)5\left(x-1\right)^5  in the Taylor Series for  f(x)=lnxf\left(x\right)=\ln x  about x = 1 is


1/5

- 1/5

1/24

- 1/24

1/120

4.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

The series n=0(9)n x4n(2n)!\sum_{n=0}^{\infty}\left(-9\right)^n\ \frac{x^{4n}}{\left(2n\right)!}  represents which function below?


 cos(3x2)\cos\left(3x^2\right)  

 sin(3x2)\sin\left(3x^2\right)  

 cos(9x4)\cos\left(9x^4\right)  

 tan1(3x)\tan^{-1}\left(3x\right)  

 e3x2e^{-3x^2}  

5.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Which of the following series converge?


n=11n\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}

n=11n\sum_{n=1}^{\infty}\frac{1}{n}

n=112n+1\sum_{n=1}^{\infty}\frac{1}{2n+1}

n=11n2+1\sum_{n=1}^{\infty}\frac{1}{n^2+1}

n=1nn2+1\sum_{n=1}^{\infty}\frac{n}{n^2+1}

6.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

Consider the series n=12n(n+1)!\sum_{n=1}^{\infty}\frac{2^n}{\left(n+1\right)!}   If the Ratio Test is applied to this series, which of the following inequalities indicates that the series is convergent?


 limn 2n+2<1\lim_{n\rightarrow\infty}\ \frac{2}{n+2}<1  

 limn n+22<1\lim_{n\rightarrow\infty}\ \frac{n+2}{2}<1  

 limn 2nn+2<1\lim_{n\rightarrow\infty}\ \frac{2^n}{n+2}<1  

 limn 2n(n+2)!<1\lim_{n\rightarrow\infty}\ \frac{2^n}{\left(n+2\right)!}<1  

7.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

What is the sum of the series

 πeπe2+πe3πe4+...\frac{\pi}{e}-\frac{\pi}{e^2}+\frac{\pi}{e^3}-\frac{\pi}{e^4}+...  

 πe+π\frac{\pi}{e+\pi}  

 πe+1\frac{\pi}{e+1}  

 πe1\frac{\pi}{e-1}  

The series diverges

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