Precalculus Examples

$\sum_{x = 5}^{10} 2^{x}$

Step 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 2

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

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Step 2.1

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

$r = \frac{2^{x + 1}}{2^{x}}$

Step 2.2

Cancel the common factor of $2^{x + 1}$ and $2^{x}$ .

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Step 2.2.1

Factor $2^{x}$ out of $2^{x + 1}$ .

$r = \frac{2^{x} \cdot 2}{2^{x}}$

Step 2.2.2

Cancel the common factors.

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Step 2.2.2.1

Multiply by $1$ .

$r = \frac{2^{x} \cdot 2}{2^{x} \cdot 1}$

Step 2.2.2.2

Cancel the common factor.

$r = \frac{2^{x} \cdot 2}{2^{x} \cdot 1}$

Step 2.2.2.3

Rewrite the expression.

$r = \frac{2}{1}$

Step 2.2.2.4

Divide $2$ by $1$ .

$r = 2$

Step 3

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

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Step 3.1

Substitute $5$ for $x$ into $2^{x}$ .

$a = 2^{5}$

Step 3.2

Raise $2$ to the power of $5$ .

$a = 32$

Step 4

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

$32 \frac{1 - 2^{6}}{1 - 1 \cdot 2}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $2$ to the power of $6$ .

$32 \frac{1 - 1 \cdot 64}{1 - 1 \cdot 2}$

Step 5.1.2

Multiply $- 1$ by $64$ .

$32 \frac{1 - 64}{1 - 1 \cdot 2}$

Step 5.1.3

Subtract $64$ from $1$ .

$32 \frac{- 63}{1 - 1 \cdot 2}$

Step 5.2

Simplify the denominator.

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Step 5.2.1

Multiply $- 1$ by $2$ .

$32 \frac{- 63}{1 - 2}$

Step 5.2.2

Subtract $2$ from $1$ .

$32 (\frac{- 63}{- 1})$

Step 5.3

Divide $- 63$ by $- 1$ .

$32 \cdot 63$

Step 5.4

Multiply $32$ by $63$ .

$2016$