Precalculus Examples

$\sum_{k = 1}^{100} 9 + {(\frac{1}{2})}^{k - 1}$

Step 1

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

$\sum_{k = 1}^{100} 9 + {(\frac{1}{2})}^{k - 1} = \sum_{k = 1}^{100} 9 + \sum_{k = 1}^{100} {(\frac{1}{2})}^{k - 1}$

Step 2

Evaluate $\sum_{k = 1}^{100} 9$ .

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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.

$(9) (100)$

Step 2.3

Multiply $9$ by $100$ .

$900$

Step 3

Evaluate $\sum_{k = 1}^{100} {(\frac{1}{2})}^{k - 1}$ .

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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 $r$ .

$r = \frac{{(\frac{1}{2})}^{k + 1 - 1}}{{(\frac{1}{2})}^{k - 1}}$

Step 3.2.2

Simplify.

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Step 3.2.2.1

Cancel the common factor of ${(\frac{1}{2})}^{k + 1 - 1}$ and ${(\frac{1}{2})}^{k - 1}$ .

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Step 3.2.2.1.1

Factor ${(\frac{1}{2})}^{k - 1}$ out of ${(\frac{1}{2})}^{k + 1 - 1}$ .

$r = \frac{{(\frac{1}{2})}^{k - 1} {(\frac{1}{2})}^{k + 0 - (k - 1)}}{{(\frac{1}{2})}^{k - 1}}$

Step 3.2.2.1.2

Cancel the common factors.

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Step 3.2.2.1.2.1

Multiply by $1$ .

$r = \frac{{(\frac{1}{2})}^{k - 1} {(\frac{1}{2})}^{k + 0 - (k - 1)}}{{(\frac{1}{2})}^{k - 1} \cdot 1}$

Step 3.2.2.1.2.2

Cancel the common factor.

$r = \frac{{(\frac{1}{2})}^{k - 1} {(\frac{1}{2})}^{k + 0 - (k - 1)}}{{(\frac{1}{2})}^{k - 1} \cdot 1}$

Step 3.2.2.1.2.3

Rewrite the expression.

$r = \frac{{(\frac{1}{2})}^{k + 0 - (k - 1)}}{1}$

Step 3.2.2.1.2.4

Divide ${(\frac{1}{2})}^{k + 0 - (k - 1)}$ by $1$ .

$r = {(\frac{1}{2})}^{k + 0 - (k - 1)}$

Step 3.2.2.2

Add $k$ and $0$ .

$r = {(\frac{1}{2})}^{k - (k - 1)}$

Step 3.2.2.3

Simplify each term.

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Step 3.2.2.3.1

Apply the distributive property.

$r = {(\frac{1}{2})}^{k - k - - 1}$

Step 3.2.2.3.2

Multiply $- 1$ by $- 1$ .

$r = {(\frac{1}{2})}^{k - k + 1}$

Step 3.2.2.4

Subtract $k$ from $k$ .

$r = {(\frac{1}{2})}^{0 + 1}$

Step 3.2.2.5

Add $0$ and $1$ .

$r = {(\frac{1}{2})}^{1}$

Step 3.2.2.6

Simplify.

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

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 ${(\frac{1}{2})}^{k - 1}$ .

$a = {(\frac{1}{2})}^{1 - 1}$

Step 3.3.2

Simplify.

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Step 3.3.2.1

Subtract $1$ from $1$ .

$a = {(\frac{1}{2})}^{0}$

Step 3.3.2.2

Apply the product rule to $\frac{1}{2}$ .

$a = \frac{1^{0}}{2^{0}}$

Step 3.3.2.3

Anything raised to $0$ is $1$ .

$a = \frac{1}{2^{0}}$

Step 3.3.2.4

Anything raised to $0$ is $1$ .

$a = \frac{1}{1}$

Step 3.3.2.5

Divide $1$ by $1$ .

$a = 1$

Step 3.4

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

$1 \frac{1 - {(\frac{1}{2})}^{100}}{1 - \frac{1}{2}}$

Step 3.5

Simplify.

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Step 3.5.1

Multiply $\frac{1 - {(\frac{1}{2})}^{100}}{1 - \frac{1}{2}}$ by $1$ .

$\frac{1 - {(\frac{1}{2})}^{100}}{1 - \frac{1}{2}}$

Step 3.5.2

Multiply the numerator and denominator of the fraction by $2$ .

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Step 3.5.2.1

Multiply $\frac{1 - {(\frac{1}{2})}^{100}}{1 - \frac{1}{2}}$ by $\frac{2}{2}$ .

$\frac{2}{2} \cdot \frac{1 - {(\frac{1}{2})}^{100}}{1 - \frac{1}{2}}$

Step 3.5.2.2

Combine.

$\frac{2 (1 - {(\frac{1}{2})}^{100})}{2 (1 - \frac{1}{2})}$

Step 3.5.3

Apply the distributive property.

$\frac{2 \cdot 1 + 2 (- {(\frac{1}{2})}^{100})}{2 \cdot 1 + 2 (- \frac{1}{2})}$

Step 3.5.4

Cancel the common factor of $2$ .

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Step 3.5.4.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$\frac{2 \cdot 1 + 2 (- {(\frac{1}{2})}^{100})}{2 \cdot 1 + 2 (\frac{- 1}{2})}$

Step 3.5.4.2

Cancel the common factor.

$\frac{2 \cdot 1 + 2 (- {(\frac{1}{2})}^{100})}{2 \cdot 1 + 2 (\frac{- 1}{2})}$

Step 3.5.4.3

Rewrite the expression.

$\frac{2 \cdot 1 + 2 (- {(\frac{1}{2})}^{100})}{2 \cdot 1 - 1}$

Step 3.5.5

Simplify the numerator.

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Step 3.5.5.1

Multiply $2$ by $1$ .

$\frac{2 + 2 (- {(\frac{1}{2})}^{100})}{2 \cdot 1 - 1}$

Step 3.5.5.2

Apply the product rule to $\frac{1}{2}$ .

$\frac{2 + 2 (- \frac{1^{100}}{2^{100}})}{2 \cdot 1 - 1}$

Step 3.5.5.3

Cancel the common factor of $2$ .

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Step 3.5.5.3.1

Move the leading negative in $- \frac{1^{100}}{2^{100}}$ into the numerator.

$\frac{2 + 2 \frac{- 1^{100}}{2^{100}}}{2 \cdot 1 - 1}$

Step 3.5.5.3.2

Factor $2$ out of $2^{100}$ .

$\frac{2 + 2 \frac{- 1^{100}}{2 \cdot 2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.3.3

Cancel the common factor.

$\frac{2 + 2 \frac{- 1^{100}}{2 \cdot 2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.3.4

Rewrite the expression.

$\frac{2 + \frac{- 1^{100}}{2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.4

One to any power is one.

$\frac{2 + \frac{- 1 \cdot 1}{2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.5

Multiply $- 1$ by $1$ .

$\frac{2 + \frac{- 1}{2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.6

Move the negative in front of the fraction.

$\frac{2 - \frac{1}{2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.7

To write $2$ as a fraction with a common denominator, multiply by $\frac{2^{99}}{2^{99}}$ .

$\frac{2 \cdot \frac{2^{99}}{2^{99}} - \frac{1}{2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.8

Combine $2$ and $\frac{2^{99}}{2^{99}}$ .

$\frac{\frac{2 \cdot 2^{99}}{2^{99}} - \frac{1}{2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.9

Combine the numerators over the common denominator.

$\frac{\frac{2 \cdot 2^{99} - 1}{2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.10

Multiply $2$ by $2^{99}$ by adding the exponents.

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Step 3.5.5.10.1

Multiply $2$ by $2^{99}$ .

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Step 3.5.5.10.1.1

Raise $2$ to the power of $1$ .

$\frac{\frac{2^{1} \cdot 2^{99} - 1}{2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.10.1.2

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

$\frac{\frac{2^{1 + 99} - 1}{2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.5.10.2

Add $1$ and $99$ .

$\frac{\frac{2^{100} - 1}{2^{99}}}{2 \cdot 1 - 1}$

Step 3.5.6

Simplify the denominator.

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Step 3.5.6.1

Multiply $2$ by $1$ .

$\frac{\frac{2^{100} - 1}{2^{99}}}{2 - 1}$

Step 3.5.6.2

Subtract $1$ from $2$ .

$\frac{\frac{2^{100} - 1}{2^{99}}}{1}$

Step 3.5.7

Divide $\frac{2^{100} - 1}{2^{99}}$ by $1$ .

$\frac{2^{100} - 1}{2^{99}}$

Step 4

Add the results of the summations.

$900 + \frac{2^{100} - 1}{2^{99}}$

Step 5

Simplify.

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Step 5.1

To write $900$ as a fraction with a common denominator, multiply by $\frac{2^{99}}{2^{99}}$ .

$900 \cdot \frac{2^{99}}{2^{99}} + \frac{2^{100} - 1}{2^{99}}$

Step 5.2

Combine $900$ and $\frac{2^{99}}{2^{99}}$ .

$\frac{900 \cdot 2^{99}}{2^{99}} + \frac{2^{100} - 1}{2^{99}}$

Step 5.3

Combine the numerators over the common denominator.

$\frac{900 \cdot 2^{99} + 2^{100} - 1}{2^{99}}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{900 \cdot 2^{99} + 2^{100} - 1}{2^{99}}$

Decimal Form:

$902$