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Precalculus : Sigma Notation

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Example Questions

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

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Example Question #1 : Sigma Notation

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

Possible Answers:

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

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

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

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

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

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:

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

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Example Question #2 : Sigma Notation

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

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

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Example Question #3 : Sigma Notation

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:

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

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Example Question #4 : 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

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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.

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

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

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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

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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

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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:

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

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Example Question #5 : Sigma Notation

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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Precalculus : Sigma Notation

Study concepts, example questions & explanations for Precalculus

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