Calculus Examples

$12$ ,

$4$ ,

$\frac{4}{3}$

Step 1

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by

$\frac{1}{3}$ gives the next term. In other words,

$a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence:

$r = \frac{1}{3}$

Step 2

The sum of a series

$S_{n}$ is calculated using the formula

$S_{n} = \frac{a (1 - r^{n})}{1 - r}$ . For the sum of an infinite geometric series

$S_{\infty}$ , as

$n$ approaches

$\infty$ ,

$1 - r^{n}$ approaches

$1$ . Thus,

$\frac{a (1 - r^{n})}{1 - r}$ approaches

$\frac{a}{1 - r}$ .

$S_{\infty} = \frac{a}{1 - r}$

Step 3

The values

$a = 12$ and

$r = \frac{1}{3}$ can be put in the equation

$S_{\infty}$ .

$S_{\infty} = \frac{12}{1 - \frac{1}{3}}$

Step 4

Simplify the equation to find

$S_{\infty}$ .

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Step 4.1

Simplify the denominator.

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Step 4.1.1

Write

$1$ as a fraction with a common denominator.

$S_{\infty} = \frac{12}{\frac{3}{3} - \frac{1}{3}}$

Step 4.1.2

Combine the numerators over the common denominator.

$S_{\infty} = \frac{12}{\frac{3 - 1}{3}}$

Step 4.1.3

Subtract

$1$ from

$3$ .

$S_{\infty} = \frac{12}{\frac{2}{3}}$

Step 4.2

Multiply the numerator by the reciprocal of the denominator.

$S_{\infty} = 12 (\frac{3}{2})$

Step 4.3

Cancel the common factor of

$2$ .

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Step 4.3.1

Factor

$2$ out of

$12$ .

$S_{\infty} = 2 (6) (\frac{3}{2})$

Step 4.3.2

Cancel the common factor.

$S_{\infty} = 2 \cdot (6 (\frac{3}{2}))$

Step 4.3.3

Rewrite the expression.

$S_{\infty} = 6 \cdot 3$

Step 4.4

Multiply

$6$ by

$3$ .

$S_{\infty} = 18$

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