Precalculus Examples

$\sum_{k = 1}^{11} \frac{1}{4} + {(- 3)}^{k - 1}$

Step 1

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

$\sum_{k = 1}^{11} \frac{1}{4} + {(- 3)}^{k - 1} = \sum_{k = 1}^{11} \frac{1}{4} + \sum_{k = 1}^{11} {(- 3)}^{k - 1}$

Step 2

Evaluate $\sum_{k = 1}^{11} \frac{1}{4}$ .

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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.

$(\frac{1}{4}) (11)$

Step 2.3

Combine $\frac{1}{4}$ and $11$ .

$\frac{11}{4}$

Step 3

Evaluate $\sum_{k = 1}^{11} {(- 3)}^{k - 1}$ .

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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 $r$ .

$r = \frac{{(- 3)}^{k + 1 - 1}}{{(- 3)}^{k - 1}}$

Step 3.2.2

Simplify.

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Step 3.2.2.1

Cancel the common factor of ${(- 3)}^{k + 1 - 1}$ and ${(- 3)}^{k - 1}$ .

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Step 3.2.2.1.1

Factor ${(- 3)}^{k - 1}$ out of ${(- 3)}^{k + 1 - 1}$ .

$r = \frac{{(- 3)}^{k - 1} {(- 3)}^{k + 0 - (k - 1)}}{{(- 3)}^{k - 1}}$

Step 3.2.2.1.2

Cancel the common factors.

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Step 3.2.2.1.2.1

Multiply by $1$ .

$r = \frac{{(- 3)}^{k - 1} {(- 3)}^{k + 0 - (k - 1)}}{{(- 3)}^{k - 1} \cdot 1}$

Step 3.2.2.1.2.2

Cancel the common factor.

$r = \frac{{(- 3)}^{k - 1} {(- 3)}^{k + 0 - (k - 1)}}{{(- 3)}^{k - 1} \cdot 1}$

Step 3.2.2.1.2.3

Rewrite the expression.

$r = \frac{{(- 3)}^{k + 0 - (k - 1)}}{1}$

Step 3.2.2.1.2.4

Divide ${(- 3)}^{k + 0 - (k - 1)}$ by $1$ .

$r = {(- 3)}^{k + 0 - (k - 1)}$

Step 3.2.2.2

Add $k$ and $0$ .

$r = {(- 3)}^{k - (k - 1)}$

Step 3.2.2.3

Simplify each term.

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Step 3.2.2.3.1

Apply the distributive property.

$r = {(- 3)}^{k - k - - 1}$

Step 3.2.2.3.2

Multiply $- 1$ by $- 1$ .

$r = {(- 3)}^{k - k + 1}$

Step 3.2.2.4

Subtract $k$ from $k$ .

$r = {(- 3)}^{0 + 1}$

Step 3.2.2.5

Add $0$ and $1$ .

$r = {(- 3)}^{1}$

Step 3.2.2.6

Evaluate the exponent.

$r = - 3$

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 $k$ into ${(- 3)}^{k - 1}$ .

$a = {(- 3)}^{1 - 1}$

Step 3.3.2

Simplify.

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Step 3.3.2.1

Subtract $1$ from $1$ .

$a = {(- 3)}^{0}$

Step 3.3.2.2

Anything raised to $0$ is $1$ .

$a = 1$

Step 3.4

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

$1 \frac{1 - {(- 3)}^{11}}{1 - - 3}$

Step 3.5

Simplify.

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Step 3.5.1

Multiply $\frac{1 - {(- 3)}^{11}}{1 - - 3}$ by $1$ .

$\frac{1 - {(- 3)}^{11}}{1 - - 3}$

Step 3.5.2

Simplify the numerator.

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Step 3.5.2.1

Raise $- 3$ to the power of $11$ .

$\frac{1 - - 177147}{1 - - 3}$

Step 3.5.2.2

Multiply $- 1$ by $- 177147$ .

$\frac{1 + 177147}{1 - - 3}$

Step 3.5.2.3

Add $1$ and $177147$ .

$\frac{177148}{1 - - 3}$

Step 3.5.3

Simplify the denominator.

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Step 3.5.3.1

Multiply $- 1$ by $- 3$ .

$\frac{177148}{1 + 3}$

Step 3.5.3.2

Add $1$ and $3$ .

$\frac{177148}{4}$

Step 3.5.4

Divide $177148$ by $4$ .

$44287$

Step 4

Add the results of the summations.

$\frac{11}{4} + 44287$

Step 5

Simplify.

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Step 5.1

To write $44287$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{11}{4} + 44287 \cdot \frac{4}{4}$

Step 5.2

Combine $44287$ and $\frac{4}{4}$ .

$\frac{11}{4} + \frac{44287 \cdot 4}{4}$

Step 5.3

Combine the numerators over the common denominator.

$\frac{11 + 44287 \cdot 4}{4}$

Step 5.4

Simplify the numerator.

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Step 5.4.1

Multiply $44287$ by $4$ .

$\frac{11 + 177148}{4}$

Step 5.4.2

Add $11$ and $177148$ .

$\frac{177159}{4}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{177159}{4}$

Decimal Form:

$44289.75$

Mixed Number Form:

$44289 \frac{3}{4}$