Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #6 : Types Of Series

Calculate the sum of the following series: $\sum^{500}_{k=1}2k+3$

Possible Answers:

252000

250000

234000

240000

Correct answer:

252000

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$ .

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

Solution:

$a_{1} = 2*1 + 3 = 5$

$a_{500} = 2 * 500 +3 = 1000 + 3 = 1003$

n =500

$\sum^{500}_{k=1}2k+3= \frac{n}{2}(a_{1}+a_{500}) = \frac{500}{2}(5+1003)=250*1008=252000$

Report an Error

Example Question #2 : Arithmetic And Geometric Series

Calculate the sum of the following series: $\sum^{300}_{k=1}3k+4$

Possible Answers:

136750

136700

136650

136600

Correct answer:

136650

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

Solution:

$a_{1} = 3*1+4 = 3+4=7$

$a_{300} = 3*300+4 = 900+4 = 904$

n = 300

$\sum^{300}_{k=1}3k+4= \frac{n}{2}(a_{1}+a_{300}) = \frac{300}{2}(7+904)=150*911=136650$

Report an Error

Example Question #1 : Arithmetic And Geometric Series

Calculate the sum of the following series: $\sum^{1200}_{k=1}6k$

Possible Answers:

4322000

4322600

4323600

4323000

Correct answer:

4323600

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

Solution:

$a_{1}=6*1=6$

$a_{1200}=6*1200 = 7200$

n =1200

$\sum^{1200}_{k=1}6k= \frac{n}{2}(a_{1}+a_{1200}) = \frac{1200}{2}(6+7200)=600*7206 = 4323600$

Report an Error

Example Question #3 : Arithmetic And Geometric Series

Calculate the sum of the following series: $\sum^{700}_{k=1}7k+5$

Possible Answers:

1720050

1720950

1720900

1720955

Correct answer:

1720950

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

Solution:

$a_{1} = 7*1+5 = 12$

$a_{700} = 7*700+5 = 4900+5 = 4905$

n = 700

$\sum^{700}_{k=1}7k+5= \frac{n}{2}(a_{1}+a_{700}) = \frac{700}{2}(12+4905) = 350*4917 = 1720950$

Report an Error

Example Question #4 : Arithmetic And Geometric Series

Calculate the sum of the following series: $\sum^{700}_{k=15}10k-4$

Possible Answers:

2449700

2447000

2449709

2449706

Correct answer:

2449706

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

When k is not equal to 1, split the sums up into 2 piece and take the difference.

Find $\sum^{n}_{k=1}10k-4 - \sum^{k-1}_{k=1}10k-4$

$a_{k}=10k-4$

$a_{1} = 10*1-4 = 6$

$a_{14} = 10*14-4 = 140-4 = 136$

$a_{700} = 700*10-4 = 6996$

$\sum^{700}_{k=1}10k-4= \frac{n}{2}(a_{1}+a_{700}) = \frac{700}{2}(6+6996) = 2450700$

$\sum^{14}_{k=1}10k-4= \frac{n}{2}(a_{14}+a_{1}) = \frac{14}{2}(136+6) = 994$

$\sum^{700}_{k=15}10k-4 = 2450700- 994=2449706$

Report an Error

Example Question #2971 : Calculus Ii

Calculate the sum of the following series: $\sum^{400}_{k=20}2k$

Possible Answers:

160000

160220

160020

160040

Correct answer:

160020

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

When k is not equal to 1, split the sums up into 2 piece and take the difference.

Find $\sum^{n}_{k=1}2k - \sum^{k-1}_{k=1}2k$

Solution:

$a_{k}=2k$

$a_{1} = 2$

$a_{19} = 2*19 = 38$

$a_{400} = 2*400 = 800$

$\sum^{400}_{k=1}2k= \frac{n}{2}(a_{1}+a_{400}) = \frac{400}{2}(2+800) = 200*802 = 160400$

$\sum^{19}_{k=1}2k= \frac{n}{2}(a_{1}+a_{19})= \frac{19}{2}(2+38) = 380$

$\sum^{400}_{k=20}2k = 160400 - 380 = 160020$

Report an Error

Example Question #2972 : Calculus Ii

Calculate the sum of the following series: $\sum^{300}_{k=100}5k-60$

Possible Answers:

188040

188940

189000

188900

Correct answer:

188940

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

When k is not equal to 1, split the sums up into 2 piece and take the difference.

Find $\sum^{n}_{k=1}5k-60 - \sum^{k-1}_{k=1}5k-60$

Solution:

$a_{k}=5k-60$

$a_{1} = 5*1-60 = -55$

$a_{99} = 5*99-60=435$

$a_{300} = 300*5-60 = 1500-60 = 1440$

$\sum^{300}_{k=1}5k-60= \frac{n}{2}(a_{1}+a_{300}) = \frac{300}{2}(-55 + 1440)=150*1385=207750$

$\sum^{99}_{k=1}5k-60= \frac{n}{2}(a_{1}+a_{99}) = \frac{99}{2}(-55+435)=\frac{99}{2}380 = 18810$

$\sum^{300}_{k=100}5k-60 = 207750 - 18810 = 188940$

Report an Error

Example Question #191 : Series In Calculus

Calculate the sum of the following series: $\sum^{100}_{k=60}15k-30$

Possible Answers:

47900

47910

47970

47930

Correct answer:

47970

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

When k is not equal to 1, split the sums up into 2 piece and take the difference.

Find $\sum^{n}_{k=1}15k-30 - \sum^{k-1}_{k=1}15k-30$

Solution:

$a_{k}=15k-30$

$a_{1}=15*1-30 = -15$

$a_{59}=15*59 -30 = 885-30 = 855$

$a_{100} = 100*15-30 = 1500-30 = 1470$

$\sum^{100}_{k=1}15k-30= \frac{n}{2}(a_{1}+a_{100})=\frac{100}{2}(-15+1470)=1455*50=72750$

$\sum^{59}_{k=1}15k-30= \frac{n}{2}(a_{59}+a_{1}) = \frac{59}{2}(855-15)=59*420=24780$

$\sum^{100}_{k=60}15k-30= 72750-24780=47970$

Report an Error

Example Question #192 : Series In Calculus

Find the value of the 700th term, or $a_{700}$ , in the following arithmetic series:

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+... =7+15+23+31+39+...$

Possible Answers:

5598

5599

6002

6000

Correct answer:

5599

Explanation:

3 pieces of information are needed to find the value of a specific term

First, find the first value, $a_{1}$ .

Second, the value of n, or the total number of terms in the series.

Finally, d, or the common difference, which can be found by calculating $a_{2} -a{1}$ .

With these pieces of information, find the value of the last with the following formula:

$a_{n} = a_{1} + (n-1)*d$

Solution:

$a_{1}=7$

n=700

$d = a_{2}-a_{1}=15-7=8$

$a_{700} = a_{1} + (n-1)*d = 7 + (700-1)*8=7+5592=5599$

Report an Error

Example Question #2975 : Calculus Ii

Find the value of the 400th term, or $a_{400}$ , in the following arithmetic series:

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+...=3+9+15+21+27+...$

Possible Answers:

2402

2399

2400

2397

Correct answer:

2397

Explanation:

3 pieces of information are needed to find the value of a specific term.

First, find the first value, . $a_{1}$

Second, the value of n, or the total number of terms in the series.

Finally, d, or the common difference, which can be found by calculating $a_{2} -a{1}$ .

With these pieces of information, find the value of the last with the following formula:

$a_{n} = a_{1} + (n-1)*d$

Solution:

$a_{1}=3$

n=400

$d = a_{2}-a_{1}=9-3=6$

$a_{400} = a_{1} + (n-1)*d=3+(400-1)*6 = 3+2394=2397$

Report an Error

View Calculus 2 Tutors

Bahaeddine
Certified Tutor

Sorbonne University, Bachelor of Science, Mathematics. Sorbonne University, Master of Science, Statistics.

View Calculus 2 Tutors

Satweka
Certified Tutor

The University of Texas at Dallas, Master of Science, Biomedical Engineering.

View Calculus 2 Tutors

Lilian Natalia
Certified Tutor

University of Texas Rio Grande Valley, Bachelor of Science, Mechanical Engineering. University of Texas Rio Grande Valley, Ma...

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Spanish Tutors in Boston, Algebra Tutors in Miami, Algebra Tutors in Boston, English Tutors in Phoenix, LSAT Tutors in New York City, SAT Tutors in Philadelphia, SSAT Tutors in Denver, ACT Tutors in Houston, Chemistry Tutors in Atlanta, SAT Tutors in New York City

Popular Courses & Classes

Spanish Courses & Classes in Philadelphia, SAT Courses & Classes in Miami, LSAT Courses & Classes in Seattle, Spanish Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Miami, ISEE Courses & Classes in New York City, SAT Courses & Classes in New York City, ACT Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Atlanta

Popular Test Prep

SSAT Test Prep in San Diego, ISEE Test Prep in New York City, SSAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Denver, LSAT Test Prep in San Francisco-Bay Area, ACT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Phoenix, SAT Test Prep in Los Angeles, LSAT Test Prep in Dallas Fort Worth, GMAT Test Prep in Dallas Fort Worth

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #6 : Types Of Series

Example Question #2 : Arithmetic And Geometric Series

Example Question #1 : Arithmetic And Geometric Series

Example Question #3 : Arithmetic And Geometric Series

Example Question #4 : Arithmetic And Geometric Series

Example Question #2971 : Calculus Ii

Example Question #2972 : Calculus Ii

Example Question #191 : Series In Calculus

Example Question #192 : Series In Calculus

Example Question #2975 : Calculus Ii

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: