Precalculus Examples

$\sum_{n = 1}^{30} - 2 + 7 (n - 1)$

Step 1

Simplify the summation.

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the distributive property.

$- 2 + 7 n + 7 \cdot - 1$

Step 1.1.2

Multiply $7$ by $- 1$ .

$- 2 + 7 n - 7$

Step 1.2

Subtract $7$ from $- 2$ .

$7 n - 9$

Step 1.3

Rewrite the summation.

$\sum_{n = 1}^{30} 7 n - 9$

Step 2

Split the summation into smaller summations that fit the summation rules.

$\sum_{n = 1}^{30} 7 n - 9 = 7 \sum_{n = 1}^{30} n + \sum_{n = 1}^{30} - 9$

Step 3

Evaluate $7 \sum_{n = 1}^{30} n$ .

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Step 3.1

The formula for the summation of a polynomial with degree $1$ is:

$\sum_{k = 1}^{n} k = \frac{n (n + 1)}{2}$

Step 3.2

Substitute the values into the formula and make sure to multiply by the front term.

$(7) (\frac{30 (30 + 1)}{2})$

Step 3.3

Simplify.

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Step 3.3.1

Add $30$ and $1$ .

$7 \frac{30 \cdot 31}{2}$

Step 3.3.2

Multiply $30$ by $31$ .

$7 (\frac{930}{2})$

Step 3.3.3

Divide $930$ by $2$ .

$7 \cdot 465$

Step 3.3.4

Multiply $7$ by $465$ .

$3255$

Step 4

Evaluate $\sum_{n = 1}^{30} - 9$ .

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Step 4.1

The formula for the summation of a constant is:

$\sum_{k = 1}^{n} c = c n$

Step 4.2

Substitute the values into the formula.

$(- 9) (30)$

Step 4.3

Multiply $- 9$ by $30$ .

$- 270$

Step 5

Add the results of the summations.

$3255 - 270$

Step 6

Subtract $270$ from $3255$ .

$2985$