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Algebra II : Mathematical Relationships and Basic Graphs

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Example Question #51 : Summations And Sequences

Solve: $\sum_{n=3}^{5}\frac{5}{n}$

Possible Answers:

$\frac{47}{12}$

$\frac{15}{n}$

$\frac{125}{27}$

$\frac{25}{3}$

$\frac{16}{3}$

Correct answer:

$\frac{47}{12}$

Explanation:

Substitute the terms n=3,4,5 for each iteration of the summation.

$\sum_{n=3}^{5}\frac{5}{n} = \frac{5}{3}+\frac{5}{4}+\frac{5}{5}$

Add the fractions.

$\frac{5}{3}+\frac{5}{4}+\frac{5}{5} = \frac{5}{3}+\frac{5}{4}+1$

Convert the fractions to a common denominator.

$\frac{20}{12}+\frac{15}{12}+\frac{12}{12} = \frac{47}{12}$

The answer is: $\frac{47}{12}$

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Example Question #52 : Summations And Sequences

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

Possible Answers:

Correct answer:

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.

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

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.

n=1; (2*1)=2

n=2; (2*2)=4

n=3; (2*3)=6

n=4; (2*4)=8

n=5; (2*5)=10

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

2+4+6+8+10=30 so.....

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

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Example Question #31 : Sigma Notation

Evaluate: $\sum_{n=0}^{3}n(n!)$

Possible Answers:

Correct answer:

Explanation:

For each iteration, we will substitute n=0,1,2,3 into the summation.

Expand the terms, and simplify the factorials.

$\sum_{n=0}^{3}n(n!)=0(0!)+1(1!)+2(2!)+3(3!)$

Zero factorial equals to one.

$= 0(1)+1(1)+2(2\times 1)+3(3\times 2\times 1)$

=0+1+4+18 = 23

The answer is:

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Example Question #32 : Sigma Notation

Solve: $\sum_{3}^{5}\frac{2}{n}$

Possible Answers:

$\frac{47}{30}$

$\frac{41}{15}$

$\frac{1}{2}$

$\frac{57}{30}$

$\frac{16}{15}$

Correct answer:

$\frac{47}{30}$

Explanation:

For each iteration, substitute n=3,4,5 into the expression of $\frac{2}{n}$ . The summation symbol indicates adding the terms together.

$\sum_{3}^{5}\frac{2}{n}=\frac{2}{3}+\frac{2}{4}+\frac{2}{5}$

Convert the fractions to a common denominator by multiplying all the denominators together.

(3)(4)(5) = 60

$\frac{2(20)}{3(20)}+\frac{2(15)}{4(15)}+\frac{2(12)}{5(12)} = \frac{40}{60}+\frac{30}{60}+\frac{24}{60}= \frac{94}{60}$

Reduce the fraction.

$\frac{94}{60}= \frac{47 \times 2}{30\times 2} = \frac{47}{30}$

The answer is: $\frac{47}{30}$

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Example Question #33 : Sigma Notation

Solve: $\sum_{n=7}^{9}(\frac{1}{2}+\frac{1}{4})$

Possible Answers:

$\frac{9}{4}$

$\frac{15}{4}$

$\frac{3}{2}$

Correct answer:

$\frac{9}{4}$

Explanation:

Simplify the fractional terms inside the parentheses.

$\sum_{n=7}^{9}(\frac{1}{2}+\frac{1}{4}) = \sum_{7}^{9}(\frac{2}{4}+\frac{1}{4}) = \sum_{7}^{9} \frac{3}{4}$

The summation starts at index seven and ends at 9. This mean that the fraction $\frac{3}{4}$ will be added to itself twice for each iteration.

$\sum_{7}^{9} \frac{3}{4} = \frac{3}{4} + \frac{3}{4} + \frac{3}{4} = \frac{9}{4}$

The answer is: $\frac{9}{4}$

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Example Question #34 : Sigma Notation

Evaluate: $\sum_{2}^{5}\frac{3}{n}$

Possible Answers:

$\frac{21}{10}$

$\frac{19}{10}$

$\frac{77}{20}$

$\frac{61}{40}$

$\frac{9}{10}$

Correct answer:

$\frac{77}{20}$

Explanation:

In order to evaluate this summation, we will need to substitute the bottom value for the first iteration, and repeat the process for each iteration until we reach to five.

Write and expand the terms.

$\sum_{2}^{5}\frac{3}{n} =\frac{3}{2} +\frac{3}{3} +\frac{3}{4} +\frac{3}{5}$

Find the least common denominator and convert the fractions to the LCD.

The least common denominator is 60.

$\frac{3}{2} +\frac{3}{3} +\frac{3}{4} +\frac{3}{5} = \frac{3(30)}{2(30)} +\frac{3(20)}{3(20)} +\frac{3(15)}{4(15)} +\frac{3(12)}{5(12)}$

Simplify the numerators.

$\frac{90}{60}+\frac{60}{60}+\frac{45}{60}+\frac{36}{60} = \frac{231}{60}$

This value can be reduced.

$\frac{231}{60} = \frac{77\times 3}{20\times 3}$

The answer is: $\frac{77}{20}$

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Example Question #51 : Mathematical Relationships And Basic Graphs

Solve: $\sum_{0}^{0}(n+2)(n+1)^n$

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

This summation will start at zero and will have no more iterations after this term.

Substitute the bottom value into the expression and solve.

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

Any value to raised to the power of zero is equal to one.

The answer is:

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Example Question #36 : Sigma Notation

Evaluate: $\sum_{n=2}^{4}-2n^2$

Possible Answers:

Correct answer:

Explanation:

Substitute n=2 for the first iteration, and then repeat for every integer until 4 is reached. Stop the loop after four.

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

Evaluate by order of operations.

=(-2(4))+(-2(9))+(-2(16)) = -8-18-32

The answer is:

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Example Question #37 : Sigma Notation

Simplify: $2\sum_{n=0}^{3}-3-n$

Possible Answers:

Correct answer:

Explanation:

The iteration starts at zero and ends at three.

Substitute the terms and increase the value of by one for each loop.

$2\sum_{n=0}^{3}-3-n = 2[(-3-0)+(-3-1)+(-3-2)+(-3-3)]$

Simplify the terms.

2[-3+(-4)+(-5)+-6] = 2[-18] = -36

The answer is:

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Example Question #38 : Sigma Notation

Evaluate

$\sum_{k = 1} ^{4} (-1) ^{k - 1} \frac{1}{k}$

Possible Answers:

$\frac{7}{12}$

$\frac{5}{12}$

$-\frac{7}{12}$

$-\frac{5}{12}$

Correct answer:

$\frac{7}{12}$

Explanation:

$\sum_{k = 1} ^{4} (-1) ^{k - 1} \frac{1}{k}$ is equal to the sum of the expressions formed by substituting 1, 2, 3, and 4, in turn, for in the expression $(-1) ^{k - 1} \frac{1}{k}$ , as follows:

k =1

$(-1) ^{k - 1} \frac{1}{k} = (-1) ^{1 - 1} \frac{1}{1} = (-1) ^{0} \cdot 1= 1 \cdot 1 = 1$

k = 2

$(-1) ^{k - 1} \frac{1}{k} = (-1) ^{2 - 1} \frac{1}{2} = (-1) ^{1} \cdot \frac{1}{2} = - 1 \cdot \frac{1}{2} = -\frac{1}{2}$

k = 3

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

k = 4

$(-1) ^{k - 1} \frac{1}{k} = (-1) ^{4 - 1} \frac{1}{4} = (-1) ^{3} \cdot \frac{1}{4} = - 1 \cdot \frac{1}{4} = - \frac{1}{4}$

The finite sequence can be restated, and evaluated, as

$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{7}{12}$ .

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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 #51 : Summations And Sequences

Example Question #52 : Summations And Sequences

Example Question #31 : Sigma Notation

Example Question #32 : Sigma Notation

Example Question #33 : Sigma Notation

Example Question #34 : Sigma Notation

Example Question #51 : Mathematical Relationships And Basic Graphs

Example Question #36 : Sigma Notation

Example Question #37 : Sigma Notation

Example Question #38 : Sigma Notation

All Algebra II Resources

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