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Calculus 2 : Types of Series

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

Example Questions

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

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

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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

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

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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

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

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

$a_{k}=10k-4$

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

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

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

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

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

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

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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:

$a_{k}=2k$

$a_{1} = 2$

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

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

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

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

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

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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:

$a_{k}=5k-60$

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

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

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

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

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

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

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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:

$a_{k}=15k-30$

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

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

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

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

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

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

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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

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

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

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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

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

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

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Example Question #2981 : Calculus Ii

Find the value of the 1200th term, or $a_{1200}$ , in the following arithmetic series:

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+...=101+124+147+170+193+...$

Possible Answers:

27668

27681

27680

27678

Correct answer:

27678

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:

$a_{1}=101$

n=1200

$d = a_{2}-a_{1}=124-101=23$

$a_{1200} = a_{1} + (n-1)*d=101+(1200-1)*23=101+27577=27678$

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Calculus 2 : Types of Series

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

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

Example Question #2981 : Calculus Ii

All Calculus 2 Resources

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