Algebra II - Summations and Sequences

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

$\frac{9}{4}$

$\frac{3}{2}$

$\frac{15}{4}$

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: $\frac{9}{4}$

Given the sequence , what is the 7th term?

Explanation

The formula for geometric sequences is defined by:

$ar^{n-1}$

The term represents the first term, while is the common ratio. The term represents the terms.

Substitute the known values.

$3(3)^{n-1}$

To determine the seventh term, simply substitute into the expression.

$3(3)^{7-1}$

The answer is:

Determine the sum, if possible: $4-\frac{2}{3}+\frac{1}{9}-...$

$\frac{24}{7}$

$\textup{The sum cannot be determined.}$

$\frac{71}{21}$

$\frac{17}{5}$

$\frac{64}{19}$

Explanation

Determine the common ratio of the infinite series by dividing the second term with the first term and the third term with the second term. The common ratios should be similar.

$\frac{-\frac{2}{3}}{4} = -\frac{2}{3} \div 4= -\frac{2}{3}\times\frac{1}{4} = -\frac{1}{6}$

$\frac{\frac{1}{9}}{-\frac{2}{3}} = \frac{1}{9} \div -\frac{2}{3} = \frac{1}{9} \times -\frac{3}{2} = -\frac{1}{6}$

Write the formula for infinite series, and substitute the terms.

$\textup{Sum} =\frac{\textup{first term}}{1-r} = \frac{4}{1-(-\frac{1}{6})} = \frac{4}{\frac{7}{6}}$

Simplify the complex fraction.

$\frac{4}{\frac{7}{6}} = 4\div \frac{7}{6} = 4 \times \frac{6}{7} = \frac{24}{7}$

The answer is: $\frac{24}{7}$

Determine the sum, if possible: $2-\frac{1}{3}+\frac{1}{18}-...$

$\frac{12}{7}$

$-\frac{1}{6}$

$\textup{The series will diverge.}$

$\frac{8}{5}$

$\frac{5}{3}$

Explanation

Write the formula for the sum of an infinite series.

$\textup{Sum} =\frac{\textup{first term}}{1-r}$

The first term is two. To determine the common ratio, we will need to divide the second term by the first, third term with the second, and so forth. The common ratio should be same for each term.

$r=\frac{-\frac{1}{3}}{2} = -\frac{1}{3}\div 2 = -\frac{1}{3} \times \frac{1}{2} = -\frac{1}{6}$

$r=\frac{\frac{1}{18}}{-\frac{1}{3}} = \frac{1}{18}\div -\frac{1}{3} = \frac{1}{18}\times -\frac{3}{1} =-\frac{1}{6}$

The common ratios are verified to be the same. Substitute the into the formula. This value must be between negative one and one or the series will diverge!

$\textup{Sum} =\frac{\textup{first term}}{1-r} = \frac{2}{1-(-\frac{1}{6})} = \frac{2}{\frac{7}{6}}$

Simplify this complex fraction.

$2\div \frac{7}{6} =2\times \frac{6}{7} =\frac{12}{7}$

The series will converge to $\frac{12}{7}$ .

Evaluate the infinite series for

Explanation

The first term of this sequence is 10. To find the common ratio r, we can just divide the second term by the first: $-9 \div 10 = - 0.9$ . So "r" is -0.9. We can find the infinite sum using the formula $\frac{a}{1-r}$ where a is the first term and r is the common ratio:

$\frac{10}{1-(-0.9)} = \frac{10}{1+0.9 } = \frac{10}{1.9} \approx 5.263$

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

Explanation

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

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

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

$\frac{9}{4}$

$\frac{3}{2}$

$\frac{15}{4}$

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: $\frac{9}{4}$

Given the sequence , what is the 7th term?

Explanation

The formula for geometric sequences is defined by:

$ar^{n-1}$

The term represents the first term, while is the common ratio. The term represents the terms.

Substitute the known values.

$3(3)^{n-1}$

To determine the seventh term, simply substitute into the expression.

$3(3)^{7-1}$

The answer is:

Evaluate: $\sum_{2}^{4}(6-2n)$

Explanation

In order to solve the summation, expand the terms of the binomial. Substitute two first, and add the quantities of each term for each integer repeating until the top integer is reached.

$\sum_{2}^{4}(6-2n)=(6-2(2))+(6-2(3))+(6-2(4))$

The answer is:

Determine the sum, if possible: $4-\frac{2}{3}+\frac{1}{9}-...$

$\frac{24}{7}$

$\textup{The sum cannot be determined.}$

$\frac{71}{21}$

$\frac{17}{5}$

$\frac{64}{19}$

Explanation

Determine the common ratio of the infinite series by dividing the second term with the first term and the third term with the second term. The common ratios should be similar.

$\frac{-\frac{2}{3}}{4} = -\frac{2}{3} \div 4= -\frac{2}{3}\times\frac{1}{4} = -\frac{1}{6}$

$\frac{\frac{1}{9}}{-\frac{2}{3}} = \frac{1}{9} \div -\frac{2}{3} = \frac{1}{9} \times -\frac{3}{2} = -\frac{1}{6}$

Write the formula for infinite series, and substitute the terms.

$\textup{Sum} =\frac{\textup{first term}}{1-r} = \frac{4}{1-(-\frac{1}{6})} = \frac{4}{\frac{7}{6}}$

Simplify the complex fraction.

$\frac{4}{\frac{7}{6}} = 4\div \frac{7}{6} = 4 \times \frac{6}{7} = \frac{24}{7}$

The answer is: $\frac{24}{7}$

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