Algebra II - Sigma Notation

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: $\frac{9}{4}$

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:

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

$\frac{77}{20}$

$\frac{21}{10}$

$\frac{9}{10}$

$\frac{19}{10}$

$\frac{61}{40}$

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: $\frac{77}{20}$

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

$15.\overline{629}$

$9.\overline{629}$

$1.\overline{185}$

$14.\overline{4}$

Explanation

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

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

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

Solve for if $x = \sum i^{2}$ for to

Explanation

For summations, we evaluate the expression at each value of , then add all of the results together.

For this problem, we are working from to .

Then adding everything up, we get

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

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

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.

The answer is:

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

Explanation

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

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

$\text{Does not exist}$

$\ln(7)$

$\ln(9)$

$\ln(2)+\ln(3)$

Explanation

The natural log domain is only valid for values greater than zero. Therefore, the solution does not exist.

Evaluate

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

$\frac{7}{12}$

$\frac{5}{12}$

$-\frac{7}{12}$

$-\frac{5}{12}$

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: