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Precalculus : Sequences and Series

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Example Questions

Example Question #21 : Sequences And Series

Solve: $\sum_{2}^{4} n+1$

Possible Answers:

Correct answer:

Explanation:

The summation starts at 2 and ends at 4. Write out the terms and solve.

$\sum_{2}^{4} n+1 = (2+1)+ (3+1)+ (4+1)= 3+4+5 =12$

The answer is:

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Example Question #21 : Sequences And Series

Write the following series in sigma notation.

4-9+14-19+...

Possible Answers:

$5\sum_{0}^{n}(-1)^{n}(4)$

$\sum_{0}^{n}(-1)^n(4+5n)$

$\sum_{0}^{n}(-1)^n(4-5n)$

$\sum_{0}^{n}(-1)^{n+1}(4+5n)$

Correct answer:

$\sum_{0}^{n}(-1)^n(4+5n)$

Explanation:

To write in sigma notation, let's make sure we have an alternating sign expression given by:

$\sum_{0}^{n}(-1)^n$

Now that we have the alternating sign, let's establish a function that increases by per term starting at . This is given by

4+5n

Putting it all together,

$\sum_{0}^{n}(-1)^n(4+5n)$

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Example Question #22 : Sequences And Series

Compute: $\sum_{2}^{5} -n+1$

Possible Answers:

Correct answer:

Explanation:

In order to solve this summation, substitute the bottom value of to the function, plus every integer until the iteration reaches to 5.

$\sum_{2}^{5} -n+1 = (-2+1 )+(-3+1) +(-4+1 )+(-5+1)$

(-1)+(-2)+(-3)+(-4) = -10

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Example Question #23 : Sequences And Series

Evaluate: $\sum_{0}^{2} 4\pi+n$

Possible Answers:

$4\pi+3$

$12\pi+3$

$12\pi+2$

$4\pi +2$

$12\pi$

Correct answer:

$12\pi+3$

Explanation:

To evaluate this, input the bottom integer into the expression $4\pi+n$ . Repeat for every integer following the bottom integer until we reach to the top integer . Sum each iteration.

$n=0 :4\pi$

$n=1: 4\pi+1$

$n=2: 4\pi+2$

Add these terms for the summation.

$\sum_{0}^{2} 4\pi+n = 4\pi + (4\pi+1)+ (4\pi +2) = 12\pi+3$

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Example Question #21 : Sequences And Series

What is the proper sigma sum notation of the summation of $1+n+n^2+...+n^{99}+n^{100}$ ?

Possible Answers:

$\sum_{j=1}^{101}j^n$

$\sum_{j=1}^{101}n^j$

$\sum_{j=100}^{0}n^j$

$\su\sum_{j=0}^{100}n^j$

Correct answer:

$\su\sum_{j=0}^{100}n^j$

Explanation:

Given that the first term of the sequence is , we know that the first term of the summation must be n^0 , and thus the lower bound of summation must be equal to There is only one option with this qualification, and so we have our answer.

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Example Question #11 : Sequences And Series

Consider the sequence: $2,\ 5,\ 8,\ 11,\ ...$

What is the fifteenth term in the sequence?

Possible Answers:

Correct answer:

Explanation:

The sequence can be described by the equation $\small 2+3(n-1)$ , where is the term in the sequence.

For the 15th term, n=15 .

$\small 2+3(n-1)=2+3(15-1)$

2+3(14)

2+42

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Example Question #1 : Terms In A Series

What is the sum of the first terms of an arithmetic series if the first term is , and the last term is ?

Possible Answers:

$\frac{181}{15}$

$\frac{271}{15}$

$\frac{317}{4}$

$\frac{315}{2}$

Correct answer:

$\frac{315}{2}$

Explanation:

Write the formula to find the arithmetic sum of a series where is the number of terms, a_1 is the first term, and a_n is the last term.

$S=\frac{n(a_1+a_n)}{2}$

Substitute the given values and solve for the sum.

$S=\frac{15(1+20)}{2}=\frac{15 \cdot21}{2}= \frac{315}{2}$

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Example Question #3 : Terms In A Series

Given the terms of the sequence $\small \small 0,1,1,2,3,5,...$ , what are the next two terms after $\small 5$ ?

Possible Answers:

$\small 7, 10$

5, 5

$\small 8, 12$

$\small 7, 13$

$\small 8, 13$

Correct answer:

$\small 8, 13$

Explanation:

The next two terms are $\small 8$ and $\small 13$ . This is the Fibonacci sequence where you start off with the terms $\small 0$ and $\small 1$ , and the next term is the sum of two previous terms. So then

$\small a_3=0+1=1$

$\small a_4=1+1=2$

$\small a_5=1+2=3$

$\small a_6 = 2+3=5$

$\small a_7=3+5=8$

$\small a_8 = 5+8=13$

$\small \small a_{9}=8+13=21$

and so on.

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Example Question #1 : Terms In A Series

What is the fifth term of the series 0.5+0.25+0.125+...

Possible Answers:

0.125125

$\frac{1}{16}$

$\frac{1}{32}$

0.1125

0.5125

Correct answer:

$\frac{1}{32}$

Explanation:

Let's try to see if this series is a geometric series.

We can divide adjacent terms to try and discover a multiplicative factor.

Doing this it seems the series proceeds with a common multiple of $\frac{1}{2}$ between each term.

Rewriting the series we get,

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\left(\frac{1}{2}\right)^n$ .

When

$n=5\rightarrow \frac{1}{2^5}=\frac{1}{32}$ .

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Example Question #1 : Terms In A Series

What is the 9th term of the series that begins 2, 4, 8, 16...

Possible Answers:

144

256

1024

512

488

Correct answer:

512

Explanation:

In this geometric series, each number is created by multiplying the previous number by 2. You may also see that, because the first number is 2, it also becomes a list of powers of 2. The list is 2, 4, 8, 16, 32, 64, 128, 256, 512, where you can see that the 9th term is 512.

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Precalculus : Sequences and Series

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #21 : Sequences And Series

Example Question #21 : Sequences And Series

Example Question #22 : Sequences And Series

Example Question #23 : Sequences And Series

Example Question #21 : Sequences And Series

Example Question #11 : Sequences And Series

Example Question #1 : Terms In A Series

Example Question #3 : Terms In A Series

Example Question #1 : Terms In A Series

Example Question #1 : Terms In A Series

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