Precalculus Examples

$81 + 27 + 9 + \dots + \frac{1}{81}$

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 $\frac{1}{3}$ gives the next term. In other words, $a_{n} = a_{1} r^{n - 1}$ .

Geometric Sequence: $r = \frac{1}{3}$

Step 2

Use the formula for a geometric sequence $a_{n} = a_{1} r^{n - 1}$ to find the number of terms, $n$ .

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Step 2.1

Substitute the values of the first term, last term, and ratio between terms into the formula.

$\frac{1}{81} = 81 {(\frac{1}{3})}^{n - 1}$

Step 2.2

Solve for $n$ .

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Step 2.2.1

Rewrite the equation as $81 {(\frac{1}{3})}^{n - 1} = \frac{1}{81}$ .

$81 {(\frac{1}{3})}^{n - 1} = \frac{1}{81}$

Step 2.2.2

Simplify $81 {(\frac{1}{3})}^{n - 1}$ .

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Step 2.2.2.1

Simplify the expression.

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Step 2.2.2.1.1

Apply the product rule to $\frac{1}{3}$ .

$81 \frac{1^{n - 1}}{3^{n - 1}} = \frac{1}{81}$

Step 2.2.2.1.2

One to any power is one.

$81 \frac{1}{3^{n - 1}} = \frac{1}{81}$

Step 2.2.2.2

Combine $81$ and $\frac{1}{3^{n - 1}}$ .

$\frac{81}{3^{n - 1}} = \frac{1}{81}$

Step 2.2.3

Multiply both sides by $3^{n - 1}$ .

$\frac{81}{3^{n - 1}} \cdot 3^{n - 1} = \frac{1}{81} \cdot 3^{n - 1}$

Step 2.2.4

Simplify.

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Step 2.2.4.1

Simplify the left side.

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Step 2.2.4.1.1

Cancel the common factor of $3^{n - 1}$ .

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Step 2.2.4.1.1.1

Cancel the common factor.

$\frac{81}{3^{n - 1}} \cdot 3^{n - 1} = \frac{1}{81} \cdot 3^{n - 1}$

Step 2.2.4.1.1.2

Rewrite the expression.

$81 = \frac{1}{81} \cdot 3^{n - 1}$

Step 2.2.4.2

Simplify the right side.

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Step 2.2.4.2.1

Combine $\frac{1}{81}$ and $3^{n - 1}$ .

$81 = \frac{3^{n - 1}}{81}$

Step 2.2.5

Solve for $n$ .

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Step 2.2.5.1

Rewrite the equation as $\frac{3^{n - 1}}{81} = 81$ .

$\frac{3^{n - 1}}{81} = 81$

Step 2.2.5.2

Multiply both sides of the equation by $81$ .

$81 \frac{3^{n - 1}}{81} = 81 \cdot 81$

Step 2.2.5.3

Simplify both sides of the equation.

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Step 2.2.5.3.1

Simplify the left side.

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Step 2.2.5.3.1.1

Cancel the common factor of $81$ .

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Step 2.2.5.3.1.1.1

Cancel the common factor.

$81 \frac{3^{n - 1}}{81} = 81 \cdot 81$

Step 2.2.5.3.1.1.2

Rewrite the expression.

$3^{n - 1} = 81 \cdot 81$

Step 2.2.5.3.2

Simplify the right side.

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Step 2.2.5.3.2.1

Multiply $81$ by $81$ .

$3^{n - 1} = 6561$

Step 2.2.5.4

Create equivalent expressions in the equation that all have equal bases.

$3^{n - 1} = 3^{8}$

Step 2.2.5.5

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$n - 1 = 8$

Step 2.2.5.6

Move all terms not containing $n$ to the right side of the equation.

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Step 2.2.5.6.1

Add $1$ to both sides of the equation.

$n = 8 + 1$

Step 2.2.5.6.2

Add $8$ and $1$ .

$n = 9$

Step 3

Use the formula for the sum of a geometric sequence $S_{n} = \frac{a_{1} (r^{n} - 1)}{r - 1}$ to find the sum.

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Step 3.1

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

$S_{n} = \frac{81 ({(\frac{1}{3})}^{9} - 1)}{\frac{1}{3} - 1}$

Step 3.2

Simplify.

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Step 3.2.1

Simplify the numerator.

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Step 3.2.1.1

Apply the product rule to $\frac{1}{3}$ .

$S_{n} = \frac{81 (\frac{1^{9}}{3^{9}} - 1)}{\frac{1}{3} - 1}$

Step 3.2.1.2

One to any power is one.

$S_{n} = \frac{81 (\frac{1}{3^{9}} - 1)}{\frac{1}{3} - 1}$

Step 3.2.1.3

Raise $3$ to the power of $9$ .

$S_{n} = \frac{81 (\frac{1}{19683} - 1)}{\frac{1}{3} - 1}$

Step 3.2.1.4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{19683}{19683}$ .

$S_{n} = \frac{81 (\frac{1}{19683} - 1 \cdot \frac{19683}{19683})}{\frac{1}{3} - 1}$

Step 3.2.1.5

Combine $- 1$ and $\frac{19683}{19683}$ .

$S_{n} = \frac{81 (\frac{1}{19683} + \frac{- 1 \cdot 19683}{19683})}{\frac{1}{3} - 1}$

Step 3.2.1.6

Combine the numerators over the common denominator.

$S_{n} = \frac{81 \frac{1 - 1 \cdot 19683}{19683}}{\frac{1}{3} - 1}$

Step 3.2.1.7

Simplify the numerator.

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Step 3.2.1.7.1

Multiply $- 1$ by $19683$ .

$S_{n} = \frac{81 \frac{1 - 19683}{19683}}{\frac{1}{3} - 1}$

Step 3.2.1.7.2

Subtract $19683$ from $1$ .

$S_{n} = \frac{81 (\frac{- 19682}{19683})}{\frac{1}{3} - 1}$

Step 3.2.1.8

Move the negative in front of the fraction.

$S_{n} = \frac{81 \cdot - 1 \frac{19682}{19683}}{\frac{1}{3} - 1}$

Step 3.2.1.9

Combine exponents.

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Step 3.2.1.9.1

Factor out negative.

$S_{n} = \frac{- (81 (\frac{19682}{19683}))}{\frac{1}{3} - 1}$

Step 3.2.1.9.2

Combine $81$ and $\frac{19682}{19683}$ .

$S_{n} = \frac{- \frac{81 \cdot 19682}{19683}}{\frac{1}{3} - 1}$

Step 3.2.1.9.3

Multiply $81$ by $19682$ .

$S_{n} = \frac{- \frac{1594242}{19683}}{\frac{1}{3} - 1}$

Step 3.2.1.10

Cancel the common factor of $1594242$ and $19683$ .

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Step 3.2.1.10.1

Factor $81$ out of $1594242$ .

$S_{n} = \frac{- \frac{81 (19682)}{19683}}{\frac{1}{3} - 1}$

Step 3.2.1.10.2

Cancel the common factors.

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Step 3.2.1.10.2.1

Factor $81$ out of $19683$ .

$S_{n} = \frac{- \frac{81 \cdot 19682}{81 \cdot 243}}{\frac{1}{3} - 1}$

Step 3.2.1.10.2.2

Cancel the common factor.

$S_{n} = \frac{- \frac{81 \cdot 19682}{81 \cdot 243}}{\frac{1}{3} - 1}$

Step 3.2.1.10.2.3

Rewrite the expression.

$S_{n} = \frac{- \frac{19682}{243}}{\frac{1}{3} - 1}$

Step 3.2.2

Simplify the denominator.

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Step 3.2.2.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$S_{n} = \frac{- \frac{19682}{243}}{\frac{1}{3} - 1 \cdot \frac{3}{3}}$

Step 3.2.2.2

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

$S_{n} = \frac{- \frac{19682}{243}}{\frac{1}{3} + \frac{- 1 \cdot 3}{3}}$

Step 3.2.2.3

Combine the numerators over the common denominator.

$S_{n} = \frac{- \frac{19682}{243}}{\frac{1 - 1 \cdot 3}{3}}$

Step 3.2.2.4

Simplify the numerator.

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Step 3.2.2.4.1

Multiply $- 1$ by $3$ .

$S_{n} = \frac{- \frac{19682}{243}}{\frac{1 - 3}{3}}$

Step 3.2.2.4.2

Subtract $3$ from $1$ .

$S_{n} = \frac{- \frac{19682}{243}}{\frac{- 2}{3}}$

Step 3.2.2.5

Move the negative in front of the fraction.

$S_{n} = \frac{- \frac{19682}{243}}{- \frac{2}{3}}$

Step 3.2.3

Dividing two negative values results in a positive value.

$S_{n} = \frac{\frac{19682}{243}}{\frac{2}{3}}$

Step 3.2.4

Multiply the numerator by the reciprocal of the denominator.

$S_{n} = \frac{19682}{243} \cdot \frac{3}{2}$

Step 3.2.5

Cancel the common factor of $2$ .

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Step 3.2.5.1

Factor $2$ out of $19682$ .

$S_{n} = \frac{2 (9841)}{243} \cdot \frac{3}{2}$

Step 3.2.5.2

Cancel the common factor.

$S_{n} = \frac{2 \cdot 9841}{243} \cdot \frac{3}{2}$

Step 3.2.5.3

Rewrite the expression.

$S_{n} = \frac{9841}{243} \cdot 3$

Step 3.2.6

Cancel the common factor of $3$ .

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Step 3.2.6.1

Factor $3$ out of $243$ .

$S_{n} = \frac{9841}{3 (81)} \cdot 3$

Step 3.2.6.2

Cancel the common factor.

$S_{n} = \frac{9841}{3 \cdot 81} \cdot 3$

Step 3.2.6.3

Rewrite the expression.

$S_{n} = \frac{9841}{81}$