Calculus Examples

$\sum_{i = 1}^{\infty} (12) {(0.25)}^{i}$

The sum of an infinite geometric series can be found using the formula $\frac{a}{1 - r}$ 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 $r = \frac{a_{i + 1}}{a_{i}}$ and simplifying.

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

$r = \frac{12 \cdot {0.25}^{i + 1}}{12 \cdot {0.25}^{i}}$

Simplify.

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Cancel the common factor of $12$ .

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Cancel the common factor.

$r = \frac{12 \cdot {0.25}^{i + 1}}{12 \cdot {0.25}^{i}}$

Rewrite the expression.

$r = \frac{{0.25}^{i + 1}}{{0.25}^{i}}$

Cancel the common factor of ${0.25}^{i + 1}$ and ${0.25}^{i}$ .

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Factor ${0.25}^{i}$ out of ${0.25}^{i + 1}$ .

$r = \frac{{0.25}^{i} \cdot 0.25}{{0.25}^{i}}$

Cancel the common factors.

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Multiply by $1$ .

$r = \frac{{0.25}^{i} \cdot 0.25}{{0.25}^{i} \cdot 1}$

Cancel the common factor.

$r = \frac{{0.25}^{i} \cdot 0.25}{{0.25}^{i} \cdot 1}$

Rewrite the expression.

$r = \frac{0.25}{1}$

Divide $0.25$ by $1$ .

$r = 0.25$

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

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Substitute $1$ for $i$ into $(12) {(0.25)}^{i}$ .

$a = 12 \cdot {0.25}^{1}$

Simplify.

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Evaluate the exponent.

$a = 12 \cdot 0.25$

Multiply $12$ by $0.25$ .

$a = 3$

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

$\frac{3}{1 - 0.25}$

Simplify.

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Subtract $0.25$ from $1$ .

$\frac{3}{0.75}$

Divide $3$ by $0.75$ .

$4$