Evaluate the Summation sum from n=1 to 8 of 3^n

$\sum_{n = 1}^{8} 3^{n}$

Step 1

The sum of a finite geometric series can be found using the formula $a (\frac{1 - r^{n}}{1 - r})$ where $a$ is the first term and $r$ is the ratio between successive terms.

Step 2

Find the ratio of successive terms by plugging into the formula $r = \frac{a_{n + 1}}{a_{n}}$ and simplifying.

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

$r = \frac{3^{n + 1}}{3^{n}}$

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

Factor $3^{n}$ out of $3^{n + 1}$ .

$r = \frac{3^{n} \cdot 3}{3^{n}}$

Cancel the common factors.

Multiply by $1$ .

$r = \frac{3^{n} \cdot 3}{3^{n} \cdot 1}$

Cancel the common factor.

$r = \frac{3^{n} \cdot 3}{3^{n} \cdot 1}$

Rewrite the expression.

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

Divide $3$ by $1$ .

$r = 3$

Step 3

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

Substitute $1$ for $n$ into $3^{n}$ .

$a = 3^{1}$

Evaluate the exponent.

$a = 3$

Step 4

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

$3 \frac{1 - 3^{8}}{1 - 1 \cdot 3}$

Step 5

Simplify.

Simplify the numerator.

Raise $3$ to the power of $8$ .

$3 \frac{1 - 1 \cdot 6561}{1 - 1 \cdot 3}$

Multiply $- 1$ by $6561$ .

$3 \frac{1 - 6561}{1 - 1 \cdot 3}$

Subtract $6561$ from $1$ .

$3 \frac{- 6560}{1 - 1 \cdot 3}$

Simplify the denominator.

Multiply $- 1$ by $3$ .

$3 \frac{- 6560}{1 - 3}$

Subtract $3$ from $1$ .

$3 (\frac{- 6560}{- 2})$

Divide $- 6560$ by $- 2$ .

$3 \cdot 3280$

Multiply $3$ by $3280$ .

$9840$

Evaluate the Summation sum from n=1 to 8 of 3^n

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