Calculus Examples

$\sum_{n = 1}^{5} 12 {(\frac{1}{2})}^{n - 1}$

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.

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

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

$r = \frac{12 {(\frac{1}{2})}^{n + 1 - 1}}{12 {(\frac{1}{2})}^{n - 1}}$

Simplify.

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Cancel the common factor of $12$ .

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Cancel the common factor.

$r = \frac{12 {(\frac{1}{2})}^{n + 1 - 1}}{12 {(\frac{1}{2})}^{n - 1}}$

Rewrite the expression.

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

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

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

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

Cancel the common factors.

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Multiply by $1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Add $n$ and $0$ .

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

Simplify each term.

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Apply the distributive property.

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

Multiply $- 1$ by $- 1$ .

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

Subtract $n$ from $n$ .

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

Add $0$ and $1$ .

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

Simplify.

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

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

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Substitute $1$ for $n$ into $12 {(\frac{1}{2})}^{n - 1}$ .

$a = 12 {(\frac{1}{2})}^{1 - 1}$

Simplify.

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Subtract $1$ from $1$ .

$a = 12 {(\frac{1}{2})}^{0}$

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

$a = 12 \frac{1^{0}}{2^{0}}$

Anything raised to $0$ is $1$ .

$a = 12 \frac{1}{2^{0}}$

Anything raised to $0$ is $1$ .

$a = 12 (\frac{1}{1})$

Cancel the common factor of $1$ .

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Cancel the common factor.

$a = 12 (\frac{1}{1})$

Rewrite the expression.

$a = 12 \cdot 1$

Multiply $12$ by $1$ .

$a = 12$

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

$12 \frac{1 - {(\frac{1}{2})}^{5}}{1 - \frac{1}{2}}$

Simplify.

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

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Multiply $\frac{1 - {(\frac{1}{2})}^{5}}{1 - \frac{1}{2}}$ by $\frac{2}{2}$ .

$12 (\frac{2}{2} \cdot \frac{1 - {(\frac{1}{2})}^{5}}{1 - \frac{1}{2}})$

Combine.

$12 \frac{2 (1 - {(\frac{1}{2})}^{5})}{2 (1 - \frac{1}{2})}$

Apply the distributive property.

$12 \frac{2 \cdot 1 + 2 (- {(\frac{1}{2})}^{5})}{2 \cdot 1 + 2 (- \frac{1}{2})}$

Cancel the common factor of $2$ .

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

$12 \frac{2 \cdot 1 + 2 (- {(\frac{1}{2})}^{5})}{2 \cdot 1 + 2 (\frac{- 1}{2})}$

Cancel the common factor.

$12 \frac{2 \cdot 1 + 2 (- {(\frac{1}{2})}^{5})}{2 \cdot 1 + 2 (\frac{- 1}{2})}$

Rewrite the expression.

$12 \frac{2 \cdot 1 + 2 (- {(\frac{1}{2})}^{5})}{2 \cdot 1 - 1}$

Simplify the numerator.

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Multiply $2$ by $1$ .

$12 \frac{2 + 2 (- {(\frac{1}{2})}^{5})}{2 \cdot 1 - 1}$

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

$12 \frac{2 + 2 (- \frac{1^{5}}{2^{5}})}{2 \cdot 1 - 1}$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1^{5}}{2^{5}}$ into the numerator.

$12 \frac{2 + 2 \frac{- 1^{5}}{2^{5}}}{2 \cdot 1 - 1}$

Factor $2$ out of $2^{5}$ .

$12 \frac{2 + 2 \frac{- 1^{5}}{2 \cdot 2^{4}}}{2 \cdot 1 - 1}$

Cancel the common factor.

$12 \frac{2 + 2 \frac{- 1^{5}}{2 \cdot 2^{4}}}{2 \cdot 1 - 1}$

Rewrite the expression.

$12 \frac{2 + \frac{- 1^{5}}{2^{4}}}{2 \cdot 1 - 1}$

One to any power is one.

$12 \frac{2 + \frac{- 1 \cdot 1}{2^{4}}}{2 \cdot 1 - 1}$

Raise $2$ to the power of $4$ .

$12 \frac{2 + \frac{- 1 \cdot 1}{16}}{2 \cdot 1 - 1}$

Multiply $- 1$ by $1$ .

$12 \frac{2 + \frac{- 1}{16}}{2 \cdot 1 - 1}$

Move the negative in front of the fraction.

$12 \frac{2 - \frac{1}{16}}{2 \cdot 1 - 1}$

To write $2$ as a fraction with a common denominator, multiply by $\frac{16}{16}$ .

$12 \frac{2 \cdot \frac{16}{16} - \frac{1}{16}}{2 \cdot 1 - 1}$

Combine $2$ and $\frac{16}{16}$ .

$12 \frac{\frac{2 \cdot 16}{16} - \frac{1}{16}}{2 \cdot 1 - 1}$

Combine the numerators over the common denominator.

$12 \frac{\frac{2 \cdot 16 - 1}{16}}{2 \cdot 1 - 1}$

Simplify the numerator.

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Multiply $2$ by $16$ .

$12 \frac{\frac{32 - 1}{16}}{2 \cdot 1 - 1}$

Subtract $1$ from $32$ .

$12 \frac{\frac{31}{16}}{2 \cdot 1 - 1}$

Simplify the denominator.

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Multiply $2$ by $1$ .

$12 \frac{\frac{31}{16}}{2 - 1}$

Subtract $1$ from $2$ .

$12 \frac{\frac{31}{16}}{1}$

Divide $\frac{31}{16}$ by $1$ .

$12 (\frac{31}{16})$

Cancel the common factor of $4$ .

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Factor $4$ out of $12$ .

$4 (3) \frac{31}{16}$

Factor $4$ out of $16$ .

$4 \cdot 3 \frac{31}{4 \cdot 4}$

Cancel the common factor.

$4 \cdot 3 \frac{31}{4 \cdot 4}$

Rewrite the expression.

$3 (\frac{31}{4})$

Combine $3$ and $\frac{31}{4}$ .

$\frac{3 \cdot 31}{4}$

Multiply $3$ by $31$ .

$\frac{93}{4}$

The result can be shown in multiple forms.

Exact Form:

$\frac{93}{4}$

Decimal Form:

$23.25$

Mixed Number Form:

$23 \frac{1}{4}$