Precalculus Examples

$\sum_{N = 1}^{8} 7 - N$

Step 1

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

$\sum_{N = 1}^{8} 7 - N = \sum_{N = 1}^{8} 7 - 1 \sum_{N = 1}^{8} N$

Step 2

Evaluate $\sum_{N = 1}^{8} 7$ .

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Step 2.1

The formula for the summation of a constant is:

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

Step 2.2

Substitute the values into the formula.

$(7) (8)$

Step 2.3

Multiply $7$ by $8$ .

$56$

Step 3

Evaluate $1 \sum_{N = 1}^{8} N$ .

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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.

$(- 1) (\frac{8 (8 + 1)}{2})$

Step 3.3

Simplify.

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Step 3.3.1

Add $8$ and $1$ .

$- 1 \frac{8 \cdot 9}{2}$

Step 3.3.2

Multiply $8$ by $9$ .

$- 1 (\frac{72}{2})$

Step 3.3.3

Divide $72$ by $2$ .

$- 1 \cdot 36$

Step 3.3.4

Multiply $- 1$ by $36$ .

$- 36$

Step 4

Add the results of the summations.

$56 - 36$

Step 5

Subtract $36$ from $56$ .

$20$