Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Sequences and Series

Study concepts, example questions & explanations for Precalculus

Create An Account

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Sequences And Series

Rewrite this sum using summation notation:

8+11+14+17+20+23+26

Possible Answers:

$\sum_{k=1}^{7} (3k)$

$\sum_{k=1}^{7} (3k+5)$

$\sum_{k=0}^{26} (3k+5)$

$\sum_{k=1}^{26} (3k+5)$

$\sum_{k=1}^{7} (k+3)$

Correct answer:

$\sum_{k=1}^{7} (3k+5)$

Explanation:

First, let's find a pattern for this sum. Each value has a difference of 3. If we know that the first value is 8, and that k will start at 1, and that each value must go up by 3, we can write the following:

8=3(k)+5

Having determined the the rule for this sum, we can now determine what value it must end at by setting the rule function equal to the last value, 26.

26 = 3k + 5

k=7

Thus the summation notation can be expressed as follows:

$\sum_{k=1}^{7} (3k+5)$

Report an Error

Example Question #1 : Sigma Notation

Write out the first 4 partial sums of the following series:

$\sum_{n=0}^{\infty }n^2sin\left(\frac{n\pi}{2}\right)$

Possible Answers:

0,1,0,-9

0,4,4,-5

1,4,9,16

0,4,0,-9

0,1,1,-8

Correct answer:

0,1,1,-8

Explanation:

Partial sums (written $s_{n}$ ) are the first few terms of a sum, so $s_3=\sum_{n=0}^{n=3}n^2sin\left(\frac{n\pi}{2}\right)=0+1+0+-9=-8$

If you then just take off the last number in that sum you get the s_2 and so on.

s_0=0=0

s_1=0+1=1

s_2=0+1+0=1

Report an Error

Example Question #1 : Sigma Notation

Express the repeating decimal 0.161616..... as a geometric series in sigma notation.

Possible Answers:

$\sum_{i=1}^{\infty}\frac{16}{100}(\frac{1}{10})^{i-1}$

$\sum_{i=0}^{\infty}\frac{4}{25}(\frac{1}{100})^{i-1}$

$\sum_{i=1}^{\infty}\frac{4}{25}(\frac{1}{100})^{i}$

$\sum_{i=1}^{\infty}\frac{4}{25}(\frac{1}{1000})^{i-1}$

$\sum_{i=1}^{\infty}\frac{4}{25}(\frac{1}{100})^{i-1}$

Correct answer:

$\sum_{i=1}^{\infty}\frac{4}{25}(\frac{1}{100})^{i-1}$

Explanation:

First break down the decimal into a sum of fractions to see the pattern.

$0.16=\frac{16}{100}, 0.0016=\frac{16}{10000}, 0.000016=\frac{16}{1000000}$

and so on. Thus,

$0.161616...=\frac{16}{100}+\frac{16}{10000}+\frac{16}{1000000}+...$

These fractions can be reduced, and the sum becomes

$\frac{4}{25}+\frac{4}{2500}+\frac{4}{250000}+...$

Each term is multiplied by $\frac{1}{100}$ to get the next term which is added.

For the first 4 terms this would look like

$\frac{4}{25}+\frac{4}{25}(\frac{1}{100})+\frac{4}{25}(\frac{1}{100})^{2}+\frac{4}{25}(\frac{1}{100})^{3}$

Let be the index variable in the sum, so if starts at the terms in the above sum would look like:

$\frac{4}{25}(\frac{1}{100})^{i-1}$ .

The decimal is repeating, so the pattern of addition occurs an infinite number of times. The sum expressed in sigma notation would then be:

$\sum_{i=1}^{\infty}\frac{4}{25}(\frac{1}{100})^{i-1}$ .

Report an Error

Example Question #12 : Sequences And Series

Evaluate the summation described by the following notation:

$\sum_{i=1}^{5}n^2-3n+10$

Possible Answers:

Correct answer:

Explanation:

In order to evaluate the summation, we must understand what the notation of the expression means:

$\sum_{i=1}^{5}n^2-3n+10$

This sigma notation tells us to sum the values obatined from evaluating the expression at each integer between and including those below and above the sigma. So we're going to start by evaluating the expression at n=1, and then add the value of the expression evaluated at n=2, and so on, until we end by adding the last value of the expression evaluated at n=5. This process is shown mathematically below:

$\sum_{i=1}^{5}n^2-3n+10=[(1)^2-3(1)+10]+...+[(5)^2-3(5)+10]$

(1)^2-3(1)+10=8

(2)^2-3(2)+10=8

(3)^2-3(3)+10=10

(4)^2-3(4)+10=14

(5)^2-3(5)+10=20

$\sum_{i=1}^{5}n^2-3n+10=8+8+10+14+20=60$

Report an Error

Example Question #11 : Sequences And Series

Evaluate: $\sum_{n=6}^{7} n+1$

Possible Answers:

Correct answer:

Explanation:

The summation starts at 6 and ends at 7. Increase the value of after each iteration:

$\sum_{n=6}^{7} n+1 = (6+1)+(7+1) =7+8 = 15$

Report an Error

Example Question #1 : Sigma Notation

Evaluate: $\sum_{n=1000}^{1001} 2-n$

Possible Answers:

Correct answer:

Explanation:

Rewrite the summation term by term and evaluate.

$\sum_{n=1000}^{1001} 2-n = (2-1000)+(2-1001)$

= -998 -999 = -1997

Report an Error

Example Question #6 : Sigma Notation

Evaluate: $\sum_{n=2}^{4} -\frac{1}{2n}$

Possible Answers:

$-\frac{1}{4}$

$-\frac{1}{9}$

$-\frac{1}{2}$

$-\frac{13}{24}$

$-\frac{1}{24}$

Correct answer:

$-\frac{13}{24}$

Explanation:

Rewrite the summation term by term:

$\sum_{n=2}^{4} -\frac{1}{2n} = \left(-\frac{1}{2\cdot 2}\right)+ \left(-\frac{1}{2\cdot 3}\right)+\left(-\frac{1}{2\cdot 4}\right)$

$=-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}$

To simplfy we get a common denominator of 24.

$=-\frac{1}{4}\cdot \frac{6}{6}-\frac{1}{6}\cdot \frac{4}{4}-\frac{1}{8}\cdot \frac{3}{3}$

$=-\frac{6}{24}-\frac{4}{24}-\frac{3}{24}=-\frac{13}{24}$

Report an Error

Example Question #5 : Sigma Notation

Evaluate:

$\sum_{w=3}^{6}(w-1)$

Possible Answers:

Correct answer:

Explanation:

$\sum_{w=3}^{6}(w-1)$ means add the values for w-1 starting with w=3 for every integer until w=6 .

This will look like:

(3-1)+(4-1)+(5-1)+(6-1) = 2 + 3 + 4 + 5 = 14

Report an Error

Example Question #6 : Sigma Notation

$-2\left[\sum_{n=1}^{5}(n-2)^2\right]$

Possible Answers:

Correct answer:

Explanation:

First, evaluate the sum. We can multiply by -2 last.

The sum

$\sum_{n=1}^{5}(n-2)^2$ means to add together every value for (n-2)^2 for an integer value of n from 1 to 5:

(1-2)^2 + (2-2)^2 + (3-2)^2 + (4-2)^2 + (5-2)^2

=(-1)^2 + 0^2 + 1^2 + 2^2 + 3^2 = 1 + 0 + 1 + 4 + 9 = 15

Now our final step is to multiply by -2.

-2*15 = -30

Report an Error

Example Question #2 : Sigma Notation

Rewrite this sum using summation notation:

$1+2+4+...+2^{28}$

Possible Answers:

$\sum_{k=0}^{29} 2^{k-1}$

$\sum_{k=1}^{29} 1+2^{k}$

$\sum_{k=1}^{29} 2^{k-1}$

$\sum_{k=1}^{29} 1-2^{k}$

$\sum_{k=1}^{28} 2^{k}$

Correct answer:

$\sum_{k=1}^{29} 2^{k-1}$

Explanation:

First, we must identify a pattern in this sum. Note that the sum can be rewritten as:

$2^0+2^1+2^2+...+2^{28}$

If we want to start our sum at k=1, then the function must be:

$2^{k-1}$ so that the first value is 2^0 .

In order to finish at $2^{28}$ , the last k value must be 29 because 29-1=28.

Thus, our summation notation is as follows:

$\sum_{k=1}^{29} 2^{k-1}$

Report an Error

View Pre-Calculus Tutors

Samantha
Certified Tutor

Middlesex County College, Associate in Science, Sustainability Studies.

View Pre-Calculus Tutors

Isaias
Certified Tutor

Instituto Tecnológico de Merida (ITM), Bachelor of Science, Biochemical Engineering. Yucatan Center for Scientific Research (...

View Pre-Calculus Tutors

Jaahanava
Certified Tutor

University of Windsor, Master's/Graduate, Computer Science .

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

SAT Tutors in Phoenix, ACT Tutors in Houston, GMAT Tutors in Denver, LSAT Tutors in San Francisco-Bay Area, MCAT Tutors in Seattle, ISEE Tutors in Seattle, MCAT Tutors in Houston, Math Tutors in Miami, MCAT Tutors in Los Angeles, GMAT Tutors in San Diego

Popular Courses & Classes

Spanish Courses & Classes in Los Angeles, Spanish Courses & Classes in San Diego, ACT Courses & Classes in San Diego, MCAT Courses & Classes in Philadelphia, GRE Courses & Classes in Seattle, MCAT Courses & Classes in Boston, GMAT Courses & Classes in Seattle, LSAT Courses & Classes in Washington DC, GRE Courses & Classes in Miami, ISEE Courses & Classes in Washington DC

Popular Test Prep

SAT Test Prep in Phoenix, SSAT Test Prep in New York City, LSAT Test Prep in Houston, MCAT Test Prep in Phoenix, ISEE Test Prep in Dallas Fort Worth, GRE Test Prep in Houston, ACT Test Prep in Atlanta, ACT Test Prep in Houston, GMAT Test Prep in Seattle, ISEE Test Prep in Los Angeles

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

Precalculus : Sequences and Series

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #11 : Sequences And Series

Example Question #1 : Sigma Notation

Example Question #1 : Sigma Notation

Example Question #12 : Sequences And Series

Example Question #11 : Sequences And Series

Example Question #1 : Sigma Notation

Example Question #6 : Sigma Notation

Example Question #5 : Sigma Notation

Example Question #6 : Sigma Notation

Example Question #2 : Sigma Notation

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: