Calculus Examples

$\sum_{i = 1}^{3} 2^{i}$

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 $r = \frac{a_{i + 1}}{a_{i}}$ and simplifying.

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

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

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

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

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

Cancel the common factors.

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Multiply by $1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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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Substitute $1$ for $i$ into $2^{i}$ .

$a = 2^{1}$

Evaluate the exponent.

$a = 2$

Step 4

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

$2 \frac{1 - 2^{3}}{1 - 1 \cdot 2}$

Step 5

Simplify.

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Simplify the numerator.

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Raise $2$ to the power of $3$ .

$2 \frac{1 - 1 \cdot 8}{1 - 1 \cdot 2}$

Multiply $- 1$ by $8$ .

$2 \frac{1 - 8}{1 - 1 \cdot 2}$

Subtract $8$ from $1$ .

$2 \frac{- 7}{1 - 1 \cdot 2}$

Simplify the denominator.

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Multiply $- 1$ by $2$ .

$2 \frac{- 7}{1 - 2}$

Subtract $2$ from $1$ .

$2 (\frac{- 7}{- 1})$

Divide $- 7$ by $- 1$ .

$2 \cdot 7$

Multiply $2$ by $7$ .

$14$