Precalculus Examples

$\sum_{i = 1}^{18} 1 + {(- 1)}^{i}$

Step 1

Split the summation into smaller summations that fit the summation rules.

$\sum_{k = 1}^{18} 1 + {(- 1)}^{i} = \sum_{k = 1}^{18} 1 + \sum_{k = 1}^{18} {(- 1)}^{i}$

Step 2

Evaluate $\sum_{k = 1}^{18} 1$ .

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Step 2.1

The formula for the summation of a constant is:

$\sum_{k = 1}^{n} c = c n$

Step 2.2

Substitute the values into the formula.

$(1) (18)$

Step 2.3

Multiply $18$ by $1$ .

$18$

Step 3

Evaluate $\sum_{k = 1}^{18} {(- 1)}^{i}$ .

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Step 3.1

The sum of a finite geometric series can be found using the formula $a (\frac{1 - r^{n}}{1 - r})$ where $a$ is the first term and $r$ is the ratio between successive terms.

Step 3.2

Find the ratio of successive terms by plugging into the formula $r = \frac{a_{k + 1}}{a_{k}}$ and simplifying.

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Step 3.2.1

Substitute $a_{k}$ and $a_{k + 1}$ into the formula for $r$ .

$r = \frac{{(- 1)}^{i + 1}}{{(- 1)}^{i}}$

Step 3.2.2

Cancel the common factor of ${(- 1)}^{i + 1}$ and ${(- 1)}^{i}$ .

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Step 3.2.2.1

Factor ${(- 1)}^{i}$ out of ${(- 1)}^{i + 1}$ .

$r = \frac{{(- 1)}^{i} \cdot - 1}{{(- 1)}^{i}}$

Step 3.2.2.2

Cancel the common factors.

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Step 3.2.2.2.1

Multiply by $1$ .

$r = \frac{{(- 1)}^{i} \cdot - 1}{{(- 1)}^{i} \cdot 1}$

Step 3.2.2.2.2

Cancel the common factor.

$r = \frac{{(- 1)}^{i} \cdot - 1}{{(- 1)}^{i} \cdot 1}$

Step 3.2.2.2.3

Rewrite the expression.

$r = \frac{- 1}{1}$

Step 3.2.2.2.4

Divide $- 1$ by $1$ .

$r = - 1$

Step 3.3

Find the first term in the series by substituting in the lower bound and simplifying.

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Step 3.3.1

Substitute $1$ for $i$ into ${(- 1)}^{i}$ .

$a = {(- 1)}^{1}$

Step 3.3.2

Evaluate the exponent.

$a = - 1$

Step 3.4

Substitute the values of the ratio, first term, and number of terms into the sum formula.

$- \frac{1 - {(- 1)}^{18}}{1 - - 1}$

Step 3.5

Simplify.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Multiply $- 1$ by ${(- 1)}^{18}$ by adding the exponents.

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Step 3.5.1.1.1

Multiply $- 1$ by ${(- 1)}^{18}$ .

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Step 3.5.1.1.1.1

Raise $- 1$ to the power of $1$ .

$- \frac{1 + {(- 1)}^{1} {(- 1)}^{18}}{1 - - 1}$

Step 3.5.1.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{1 + {(- 1)}^{1 + 18}}{1 - - 1}$

Step 3.5.1.1.2

Add $1$ and $18$ .

$- \frac{1 + {(- 1)}^{19}}{1 - - 1}$

Step 3.5.1.2

Raise $- 1$ to the power of $19$ .

$- \frac{1 - 1}{1 - - 1}$

Step 3.5.1.3

Subtract $1$ from $1$ .

$- \frac{0}{1 - - 1}$

Step 3.5.2

Simplify the denominator.

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Step 3.5.2.1

Multiply $- 1$ by $- 1$ .

$- \frac{0}{1 + 1}$

Step 3.5.2.2

Add $1$ and $1$ .

$- \frac{0}{2}$

Step 3.5.3

Divide $0$ by $2$ .

$- 0$

Step 3.5.4

Multiply $- 1$ by $0$ .

$0$

Step 4

Add the results of the summations.

$18 + 0$

Step 5

Add $18$ and $0$ .

$18$