Trigonometry Examples

1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \frac{1}{81}

$1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \frac{1}{81}$

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}

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

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

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

Geometric Sequence:

r = - \frac{1}{3}

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

Step 2

This is the form of a geometric sequence.

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

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

Step 3

Substitute in the values of

a_{1} = 1

$a_{1} = 1$ and

r = - \frac{1}{3}

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

a_{n} = 1 {(- \frac{1}{3})}^{n - 1}

$a_{n} = 1 {(- \frac{1}{3})}^{n - 1}$

Step 4

Multiply

{(- \frac{1}{3})}^{n - 1}

${(- \frac{1}{3})}^{n - 1}$ by

1

$1$ .

a_{n} = {(- \frac{1}{3})}^{n - 1}

$a_{n} = {(- \frac{1}{3})}^{n - 1}$

Step 5

Use the power rule

{(a b)}^{n} = a^{n} b^{n}

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

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Step 5.1

Apply the product rule to

- \frac{1}{3}

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

a_{n} = {(- 1)}^{n - 1} {(\frac{1}{3})}^{n - 1}

$a_{n} = {(- 1)}^{n - 1} {(\frac{1}{3})}^{n - 1}$

Step 5.2

Apply the product rule to

\frac{1}{3}

$\frac{1}{3}$ .

a_{n} = {(- 1)}^{n - 1} \frac{1^{n - 1}}{3^{n - 1}}

$a_{n} = {(- 1)}^{n - 1} \frac{1^{n - 1}}{3^{n - 1}}$

a_{n} = {(- 1)}^{n - 1} \frac{1^{n - 1}}{3^{n - 1}}

$a_{n} = {(- 1)}^{n - 1} \frac{1^{n - 1}}{3^{n - 1}}$

Step 6

One to any power is one.

a_{n} = {(- 1)}^{n - 1} \frac{1}{3^{n - 1}}

$a_{n} = {(- 1)}^{n - 1} \frac{1}{3^{n - 1}}$

Step 7

Combine

{(- 1)}^{n - 1}

${(- 1)}^{n - 1}$ and

\frac{1}{3^{n - 1}}

$\frac{1}{3^{n - 1}}$ .

a_{n} = \frac{{(- 1)}^{n - 1}}{3^{n - 1}}

$a_{n} = \frac{{(- 1)}^{n - 1}}{3^{n - 1}}$

Step 8

This is the formula to find the sum of the first

n

$n$ terms of the geometric sequence. To evaluate it, find the values of

r

$r$ and

a_{1}

$a_{1}$ .

S_{n} = \frac{a_{1} (r^{n} - 1)}{r - 1}

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

Step 9

Replace the variables with the known values to find

S_{5}

$S_{5}$ .

S_{5} = 1 \cdot \frac{{(- \frac{1}{3})}^{5} - 1}{- \frac{1}{3} - 1}

$S_{5} = 1 \cdot \frac{{(- \frac{1}{3})}^{5} - 1}{- \frac{1}{3} - 1}$

Step 10

Multiply

\frac{{(- \frac{1}{3})}^{5} - 1}{- \frac{1}{3} - 1}

$\frac{{(- \frac{1}{3})}^{5} - 1}{- \frac{1}{3} - 1}$ by

1

$1$ .

S_{5} = \frac{{(- \frac{1}{3})}^{5} - 1}{- \frac{1}{3} - 1}

$S_{5} = \frac{{(- \frac{1}{3})}^{5} - 1}{- \frac{1}{3} - 1}$

Step 11

Multiply the numerator and denominator of the fraction by

3

$3$ .

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Step 11.1

Multiply

\frac{{(- \frac{1}{3})}^{5} - 1}{- \frac{1}{3} - 1}

$\frac{{(- \frac{1}{3})}^{5} - 1}{- \frac{1}{3} - 1}$ by

\frac{3}{3}

$\frac{3}{3}$ .

S_{5} = \frac{3}{3} \cdot \frac{{(- \frac{1}{3})}^{5} - 1}{- \frac{1}{3} - 1}

$S_{5} = \frac{3}{3} \cdot \frac{{(- \frac{1}{3})}^{5} - 1}{- \frac{1}{3} - 1}$

Step 11.2

Combine.

S_{5} = \frac{3 ({(- \frac{1}{3})}^{5} - 1)}{3 (- \frac{1}{3} - 1)}

$S_{5} = \frac{3 ({(- \frac{1}{3})}^{5} - 1)}{3 (- \frac{1}{3} - 1)}$

S_{5} = \frac{3 ({(- \frac{1}{3})}^{5} - 1)}{3 (- \frac{1}{3} - 1)}

$S_{5} = \frac{3 ({(- \frac{1}{3})}^{5} - 1)}{3 (- \frac{1}{3} - 1)}$

Step 12

Apply the distributive property.

S_{5} = \frac{3 {(- \frac{1}{3})}^{5} + 3 \cdot - 1}{3 (- \frac{1}{3}) + 3 \cdot - 1}

$S_{5} = \frac{3 {(- \frac{1}{3})}^{5} + 3 \cdot - 1}{3 (- \frac{1}{3}) + 3 \cdot - 1}$

Step 13

Cancel the common factor of

3

$3$ .

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Step 13.1

Move the leading negative in

- \frac{1}{3}

$- \frac{1}{3}$ into the numerator.

S_{5} = \frac{3 {(- \frac{1}{3})}^{5} + 3 \cdot - 1}{3 (\frac{- 1}{3}) + 3 \cdot - 1}

$S_{5} = \frac{3 {(- \frac{1}{3})}^{5} + 3 \cdot - 1}{3 (\frac{- 1}{3}) + 3 \cdot - 1}$

Step 13.2

Cancel the common factor.

$S_{5} = \frac{3 {(- \frac{1}{3})}^{5} + 3 \cdot - 1}{3 (\frac{- 1}{3}) + 3 \cdot - 1}$

Step 13.3

Rewrite the expression.

$S_{5} = \frac{3 {(- \frac{1}{3})}^{5} + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14

Simplify the numerator.

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Step 14.1

Use the power rule

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

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Step 14.1.1

Apply the product rule to

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

$S_{5} = \frac{3 ({(- 1)}^{5} {(\frac{1}{3})}^{5}) + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.1.2

Apply the product rule to

$\frac{1}{3}$ .

$S_{5} = \frac{3 ({(- 1)}^{5} (\frac{1^{5}}{3^{5}})) + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.2

Raise

$- 1$ to the power of

$5$ .

$S_{5} = \frac{3 (- \frac{1^{5}}{3^{5}}) + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.3

Cancel the common factor of

$3$ .

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Step 14.3.1

Move the leading negative in

$- \frac{1^{5}}{3^{5}}$ into the numerator.

$S_{5} = \frac{3 (\frac{- 1^{5}}{3^{5}}) + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.3.2

Factor

$3$ out of

$3^{5}$ .

$S_{5} = \frac{3 (\frac{- 1^{5}}{3 \cdot 3^{4}}) + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.3.3

Cancel the common factor.

$S_{5} = \frac{3 (\frac{- 1^{5}}{3 \cdot 3^{4}}) + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.3.4

Rewrite the expression.

$S_{5} = \frac{\frac{- 1^{5}}{3^{4}} + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.4

One to any power is one.

$S_{5} = \frac{\frac{- 1 \cdot 1}{3^{4}} + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.5

Raise

$3$ to the power of

$4$ .

$S_{5} = \frac{\frac{- 1 \cdot 1}{81} + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.6

Multiply

$- 1$ by

$1$ .

$S_{5} = \frac{\frac{- 1}{81} + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.7

Move the negative in front of the fraction.

$S_{5} = \frac{- \frac{1}{81} + 3 \cdot - 1}{- 1 + 3 \cdot - 1}$

Step 14.8

Multiply

$3$ by

$- 1$ .

$S_{5} = \frac{- \frac{1}{81} - 3}{- 1 + 3 \cdot - 1}$

Step 14.9

To write

$- 3$ as a fraction with a common denominator, multiply by

$\frac{81}{81}$ .

$S_{5} = \frac{- \frac{1}{81} - 3 \cdot \frac{81}{81}}{- 1 + 3 \cdot - 1}$

Step 14.10

Combine

$- 3$ and

$\frac{81}{81}$ .

$S_{5} = \frac{- \frac{1}{81} + \frac{- 3 \cdot 81}{81}}{- 1 + 3 \cdot - 1}$

Step 14.11

Combine the numerators over the common denominator.

$S_{5} = \frac{\frac{- 1 - 3 \cdot 81}{81}}{- 1 + 3 \cdot - 1}$

Step 14.12

Simplify the numerator.

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Step 14.12.1

Multiply

$- 3$ by

$81$ .

$S_{5} = \frac{\frac{- 1 - 243}{81}}{- 1 + 3 \cdot - 1}$

Step 14.12.2

Subtract

$243$ from

$- 1$ .

$S_{5} = \frac{\frac{- 244}{81}}{- 1 + 3 \cdot - 1}$

Step 14.13

Move the negative in front of the fraction.

$S_{5} = \frac{- \frac{244}{81}}{- 1 + 3 \cdot - 1}$

Step 15

Simplify the denominator.

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Step 15.1

Multiply

$3$ by

$- 1$ .

$S_{5} = \frac{- \frac{244}{81}}{- 1 - 3}$

Step 15.2

Subtract

$3$ from

$- 1$ .

$S_{5} = \frac{- \frac{244}{81}}{- 4}$

Step 16

Multiply the numerator by the reciprocal of the denominator.

$S_{5} = - \frac{244}{81} \cdot \frac{1}{- 4}$

Step 17

Cancel the common factor of

$4$ .

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Step 17.1

Move the leading negative in

$- \frac{244}{81}$ into the numerator.

$S_{5} = \frac{- 244}{81} \cdot \frac{1}{- 4}$

Step 17.2

Factor

$4$ out of

$- 244$ .

$S_{5} = \frac{4 (- 61)}{81} \cdot \frac{1}{- 4}$

Step 17.3

Factor

$4$ out of

$- 4$ .

$S_{5} = \frac{4 \cdot - 61}{81} \cdot \frac{1}{4 \cdot - 1}$

Step 17.4

Cancel the common factor.

$S_{5} = \frac{4 \cdot - 61}{81} \cdot \frac{1}{4 \cdot - 1}$

Step 17.5

Rewrite the expression.

$S_{5} = \frac{- 61}{81} \cdot \frac{1}{- 1}$

Step 18

Multiply

$\frac{- 61}{81}$ by

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

$S_{5} = \frac{- 61}{81 \cdot - 1}$

Step 19

Multiply

$81$ by

$- 1$ .

$S_{5} = \frac{- 61}{- 81}$

Step 20

Dividing two negative values results in a positive value.

$S_{5} = \frac{61}{81}$