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Algebra II : Mathematical Relationships and Basic Graphs

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Example Questions

Example Question #41 : Mathematical Relationships And Basic Graphs

Evaluate the summation: $\sum_{n=3}^{6} (1-6n)$

Possible Answers:

Correct answer:

Explanation:

This summation will repeat three times, for n=4,5,6 .

Substitute the first term into the parentheses, repeat, and sum the process for the next three terms.

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

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

Simplify each term.

(-17)+(-23)+(-29)+(-35) =-104

The answer is:

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Example Question #42 : Mathematical Relationships And Basic Graphs

Determine the value of: $\sum_{0}^{2}2n^6$

Possible Answers:

128

130

$\textup{The answer is not given.}$

Correct answer:

130

Explanation:

Expand the summation sign. Start with zero as the first index, then one, and finally two. Since two is the top digit, the summation will stop.

$\sum_{0}^{2}2n^6=2(0)^6+2(1)^6+2(2)^6$

Simplify these terms.

0+2+2(64) = 2+128 = 130

The answer is: 130

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Example Question #41 : Mathematical Relationships And Basic Graphs

$\sum_{k=1}^{4} k^2 - 7$

Possible Answers:

104

Correct answer:

Explanation:

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

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

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

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Example Question #22 : Sigma Notation

$\sum_{n=0}^{4} \binom{4}{n}$

Possible Answers:

$8.\overline{3}$

Correct answer:

Explanation:

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

$\binom{4}{0 } + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4}$

$\frac{4!}{0! \cdot 4!} + \frac{4!}{1!\cdot3! } + \frac{4!}{2! \cdot 2! } + \frac{4! } { 3! \cdot 1! } + \frac{4!}{4! \cdot 0!}$

1 + 4 + 6 + 4 + 1 = 16

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Example Question #2703 : Algebra Ii

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

Possible Answers:

17.5

Correct answer:

17.5

Explanation:

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

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

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

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

10 + 5 + 2.5 = 17.5

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Example Question #42 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

Correct answer:

Explanation:

This is asking us to add n-5 for n = 3 and 4.

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

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Example Question #43 : Mathematical Relationships And Basic Graphs

Calculate $\sum_{n=0}^{5} \frac{1}{2}n^2 +1$

Possible Answers:

32.5

33.5

27.5

Correct answer:

33.5

Explanation:

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

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

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

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Example Question #44 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

$1.\overline{185}$

$14.\overline{4}$

$15.\overline{629}$

$9.\overline{629}$

Correct answer:

$15.\overline{629}$

Explanation:

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

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

$= 6 + 4 + 2.\overline{6} + 1. \overline{7} + 1.\overline{185} = 15. \overline{629}$

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Example Question #45 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

330

255

285

Correct answer:

255

Explanation:

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

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

=25 + 36+49+64+81 = 255

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Example Question #46 : Mathematical Relationships And Basic Graphs

Evaluate $\frac{1}{3} \sum_{k=0}^{4} 2k-1$

Possible Answers:

$\frac{17}{3}$

Correct answer:

Explanation:

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

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

= -1 + 1 + 3 + 5 + 7 = 15

Now, multiply by $\frac{1}{3} : \frac{1}{3} \cdot 15 = 5$

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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #41 : Mathematical Relationships And Basic Graphs

Example Question #42 : Mathematical Relationships And Basic Graphs

Example Question #41 : Mathematical Relationships And Basic Graphs

Example Question #22 : Sigma Notation

Example Question #2703 : Algebra Ii

Example Question #42 : Mathematical Relationships And Basic Graphs

Example Question #43 : Mathematical Relationships And Basic Graphs

Example Question #44 : Mathematical Relationships And Basic Graphs

Example Question #45 : Mathematical Relationships And Basic Graphs

Example Question #46 : Mathematical Relationships And Basic Graphs

All Algebra II Resources

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