Precalculus Examples

$1 - \frac{1}{2}$ , $\frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \frac{1}{32} + \frac{1}{64}$

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 $- \frac{1}{2}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = - \frac{1}{2}$

Step 2

This is the form of a geometric sequence.

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

Step 3

Substitute in the values of $a_{1} = 1$ and $r = - \frac{1}{2}$ .

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

Step 4

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

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

Step 5

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 5.1

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

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

Step 5.2

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

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

Step 6

One to any power is one.

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

Step 7

Combine ${(- 1)}^{n - 1}$ and $\frac{1}{2^{n - 1}}$ .

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

Step 8

Substitute in the value of $n$ to find the $n$ th term.

$a_{8} = \frac{{(- 1)}^{(8) - 1}}{2^{(8) - 1}}$

Step 9

Simplify the numerator.

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Step 9.1

Subtract $1$ from $8$ .

$a_{8} = \frac{{(- 1)}^{7}}{2^{8 - 1}}$

Step 9.2

Raise $- 1$ to the power of $7$ .

$a_{8} = \frac{- 1}{2^{8 - 1}}$

Step 10

Simplify the denominator.

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Step 10.1

Subtract $1$ from $8$ .

$a_{8} = \frac{- 1}{2^{7}}$

Step 10.2

Raise $2$ to the power of $7$ .

$a_{8} = \frac{- 1}{128}$

Step 11

Move the negative in front of the fraction.

$a_{8} = - \frac{1}{128}$