Calculus Examples

$\sum_{i = 4}^{7} 3^{i}$

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 $r = \frac{a_{i + 1}}{a_{i}}$ and simplifying.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Cancel the common factors.

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Step 2.2.2.1

Multiply by $1$ .

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

Step 2.2.2.2

Cancel the common factor.

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

Step 2.2.2.3

Rewrite the expression.

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

Step 2.2.2.4

Divide $3$ by $1$ .

$r = 3$

Step 3

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

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Step 3.1

Substitute $4$ for $i$ into $3^{i}$ .

$a = 3^{4}$

Step 3.2

Raise $3$ to the power of $4$ .

$a = 81$

Step 4

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

$81 \frac{1 - 3^{4}}{1 - 1 \cdot 3}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $3$ to the power of $4$ .

$81 \frac{1 - 1 \cdot 81}{1 - 1 \cdot 3}$

Step 5.1.2

Multiply $- 1$ by $81$ .

$81 \frac{1 - 81}{1 - 1 \cdot 3}$

Step 5.1.3

Subtract $81$ from $1$ .

$81 \frac{- 80}{1 - 1 \cdot 3}$

Step 5.2

Simplify the denominator.

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Step 5.2.1

Multiply $- 1$ by $3$ .

$81 \frac{- 80}{1 - 3}$

Step 5.2.2

Subtract $3$ from $1$ .

$81 (\frac{- 80}{- 2})$

Step 5.3

Divide $- 80$ by $- 2$ .

$81 \cdot 40$

Step 5.4

Multiply $81$ by $40$ .

$3240$