Algebra II : Sigma Notation

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #21 : Sigma Notation

\(\displaystyle \sum_{k=1}^{4} k^2 - 7\)

Possible Answers:

\(\displaystyle -7\)

\(\displaystyle 2\)

\(\displaystyle 104\)

\(\displaystyle 27\)

Correct answer:

\(\displaystyle 2\)

Explanation:

This notation is asking us to add \(\displaystyle k^2 - 7\) for all integer values of k between 1 and 4: 

\(\displaystyle (1^2 - 7 ) + (2^2 - 7 ) + (3^2 - 7 ) + (4^2 - 7 )\)

\(\displaystyle (1-7) + (4-7) + (9-7) + (16-7 ) = -6 + -3 + 2 + 9 = 2\)

 

Example Question #42 : Mathematical Relationships And Basic Graphs

\(\displaystyle \sum_{n=0}^{4} \binom{4}{n}\)

Possible Answers:

\(\displaystyle 14\)

\(\displaystyle 8.\overline{3}\)

\(\displaystyle 15\)

\(\displaystyle 16\)

Correct answer:

\(\displaystyle 16\)

Explanation:

This expression is asking us to add the expressions \(\displaystyle \binom{4}{n}\) for every integer value of n from 0 to 4:

\(\displaystyle \binom{4}{0 } + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4}\)

\(\displaystyle \frac{4!}{0! \cdot 4!} + \frac{4!}{1!\cdot3! } + \frac{4!}{2! \cdot 2! } + \frac{4! } { 3! \cdot 1! } + \frac{4!}{4! \cdot 0!}\)

\(\displaystyle 1 + 4 + 6 + 4 + 1 = 16\)

Example Question #43 : Mathematical Relationships And Basic Graphs

\(\displaystyle \sum_{k=1}^{3} 20 \cdot (\frac{1}{2} ) ^k\)

Possible Answers:

\(\displaystyle 17.5\)

\(\displaystyle 15\)

\(\displaystyle 30\)

\(\displaystyle 35\)

Correct answer:

\(\displaystyle 17.5\)

Explanation:

\(\displaystyle \sum_{k=1}^{3} 20 \cdot (\frac{1}{2} ) ^k\)means that we are adding half of 20, plus half of that, plus half of that.

1: \(\displaystyle 20 \cdot (\frac{1}{2} ) = 10\)

2: \(\displaystyle 10(\frac{1}{2} ) = 5\)

3: \(\displaystyle 5(\frac{1}{2} ) = 2.5\)

\(\displaystyle 10 + 5 + 2.5 = 17.5\)

Example Question #22 : Sigma Notation

\(\displaystyle \sum_{n=3}^{5} \left (n-5 \right )\)

Possible Answers:

\(\displaystyle -9\)

\(\displaystyle -3\)

\(\displaystyle 2\)

\(\displaystyle 3\)

Correct answer:

\(\displaystyle -3\)

Explanation:

This is asking us to add \(\displaystyle n-5\) for n = 3 and 4.

\(\displaystyle (3 -5 ) + (4 - 5 ) = -2 + -1 = -3\)

Example Question #23 : Sigma Notation

Calculate \(\displaystyle \sum_{n=0}^{5} \frac{1}{2}n^2 +1\)

Possible Answers:

\(\displaystyle 33.5\)

\(\displaystyle 20\)

\(\displaystyle 32.5\)

\(\displaystyle 27.5\)

Correct answer:

\(\displaystyle 33.5\)

Explanation:

This is asking us to plug in the integers between 0 and 5, then add these numbers together.

\(\displaystyle (\frac{1}{2} (0)^2 +1 ) + (\frac{1}{2} (1)^2 +1 ) + (\frac{1}{2} (2)^2 +1 ) + (\frac{1}{2} (3)^2 +1 ) + (\frac{1}{2} (4)^2 +1 ) +(\frac{1}{2} (5)^2 +1 )\)

\(\displaystyle = (0+1) + (.5+1) + (2+1) + (4.5+1) + (8+1) + (12.5+1) = 33.5\)

Example Question #24 : Sigma Notation

Evaluate \(\displaystyle \sum_{n=0}^{4} 6 \cdot(\frac{2}{3})^n\)

Possible Answers:

\(\displaystyle 15.\overline{629}\)

\(\displaystyle 9.\overline{629}\)

\(\displaystyle 14.\overline{4}\)

\(\displaystyle 1.\overline{185}\)

Correct answer:

\(\displaystyle 15.\overline{629}\)

Explanation:

This is asking us to add 6 plus two thirds of 6, plus two thirds of that, etc.

\(\displaystyle 6 + 6 \cdot \frac{2}{3} + 6 \cdot\left ( \frac{2}{3} \right )^2 + 6 \cdot \left (\frac{2}{3} \right ) ^3 + 6 \cdot \left (\frac{2}{3} \right )^4\)

\(\displaystyle = 6 + 4 + 2.\overline{6} + 1. \overline{7} + 1.\overline{185} = 15. \overline{629}\)

Example Question #25 : Sigma Notation

Evaluate \(\displaystyle \sum_{n=4}^{8} (n+1)^2\)

Possible Answers:

\(\displaystyle 54\)

\(\displaystyle 330\)

\(\displaystyle 285\)

\(\displaystyle 255\)

Correct answer:

\(\displaystyle 255\)

Explanation:

This is asking us to substitute integer values between 4 and 8 for n, and then add the results.

\(\displaystyle (4+1)^2 + (5+1)^2 + (6+1)^2 + (7+1)^2 + (8+1)^2\)

\(\displaystyle =25 + 36+49+64+81 = 255\)

Example Question #48 : Mathematical Relationships And Basic Graphs

Evaluate \(\displaystyle \frac{1}{3} \sum_{k=0}^{4} 2k-1\)

Possible Answers:

\(\displaystyle 5\)

\(\displaystyle \frac{17}{3}\)

\(\displaystyle 3\)

\(\displaystyle -17\)

Correct answer:

\(\displaystyle 5\)

Explanation:

First, evaluate the sigma expression. It is asking us to plug in for k all of the integers between 0 and 4:

\(\displaystyle \left [2(0)-1 \right ] + \left [2(1)-1 \right ] + \left [2(2) - 1 \right ]+ \left [2(3) - 1 \right ] + \left [2(4) -1 \right ]\)

\(\displaystyle = -1 + 1 + 3 + 5 + 7 = 15\)

Now, multiply by \(\displaystyle \frac{1}{3} : \frac{1}{3} \cdot 15 = 5\)

Example Question #51 : Summations And Sequences

Solve:  \(\displaystyle \sum_{n=3}^{5}\frac{5}{n}\)

Possible Answers:

\(\displaystyle \frac{15}{n}\)

\(\displaystyle \frac{25}{3}\)

\(\displaystyle \frac{125}{27}\)

\(\displaystyle \frac{47}{12}\)

\(\displaystyle \frac{16}{3}\)

Correct answer:

\(\displaystyle \frac{47}{12}\)

Explanation:

Substitute the terms \(\displaystyle n=3,4,5\) for each iteration of the summation.

\(\displaystyle \sum_{n=3}^{5}\frac{5}{n} = \frac{5}{3}+\frac{5}{4}+\frac{5}{5}\)

Add the fractions.

\(\displaystyle \frac{5}{3}+\frac{5}{4}+\frac{5}{5} = \frac{5}{3}+\frac{5}{4}+1\)

Convert the fractions to a common denominator. 

\(\displaystyle \frac{20}{12}+\frac{15}{12}+\frac{12}{12} = \frac{47}{12}\)

The answer is:  \(\displaystyle \frac{47}{12}\)

Example Question #52 : Summations And Sequences

\(\displaystyle \sum^5_{n=1}(2n)=?\)

Possible Answers:

\(\displaystyle 5\)

\(\displaystyle 10\)

\(\displaystyle 30\)

\(\displaystyle 15\)

Correct answer:

\(\displaystyle 30\)

Explanation:

When using summation notation, the number on the bottom is the starting input value, and the number on the top is the ending input value.

 \(\displaystyle \sum^5_{n=1}(2n)=?\)

This means that you will be adding the result of the first 5 outputs after plugging in 1 through 5. 

Plug in each value of n, from 1 to 5.

\(\displaystyle n=1; (2*1)=2\)

\(\displaystyle n=2; (2*2)=4\)

\(\displaystyle n=3; (2*3)=6\)

\(\displaystyle n=4; (2*4)=8\)

\(\displaystyle n=5; (2*5)=10\)

Now we take all of these numbers and sum them to get: 

\(\displaystyle 2+4+6+8+10=30\) so.....

\(\displaystyle \sum^5_{n=1}(2n)=30\)

 

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