Precalculus Examples

$\sum_{x = 1}^{90} 8 - \frac{1}{3} x$

Step 1

Simplify the summation.

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Step 1.1

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

$8 - \frac{x}{3}$

Step 1.2

Rewrite the summation.

$\sum_{x = 1}^{90} 8 - \frac{x}{3}$

Step 2

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

$\sum_{x = 1}^{90} 8 - \frac{x}{3} = \sum_{x = 1}^{90} 8 + \sum_{x = 1}^{90} - \frac{x}{3}$

Step 3

Evaluate $\sum_{x = 1}^{90} 8$ .

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Step 3.1

The formula for the summation of a constant is:

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

Step 3.2

Substitute the values into the formula.

$(8) (90)$

Step 3.3

Multiply $8$ by $90$ .

$720$

Step 4

Evaluate $\sum_{x = 1}^{90} - \frac{x}{3}$ .

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Step 4.1

Factor $- \frac{1}{3}$ out of the summation.

$(- \frac{1}{3}) \sum_{x = 1}^{90} x$

Step 4.2

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

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

Step 4.3

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

$(- \frac{1}{3}) (\frac{90 (90 + 1)}{2})$

Step 4.4

Simplify.

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Step 4.4.1

Simplify the expression.

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Step 4.4.1.1

Add $90$ and $1$ .

$- \frac{1}{3} \cdot \frac{90 \cdot 91}{2}$

Step 4.4.1.2

Multiply $90$ by $91$ .

$- \frac{1}{3} \cdot \frac{8190}{2}$

Step 4.4.2

Cancel the common factor of $3$ .

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Step 4.4.2.1

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$\frac{- 1}{3} \cdot \frac{8190}{2}$

Step 4.4.2.2

Factor $3$ out of $8190$ .

$\frac{- 1}{3} \cdot \frac{3 (2730)}{2}$

Step 4.4.2.3

Cancel the common factor.

$\frac{- 1}{3} \cdot \frac{3 \cdot 2730}{2}$

Step 4.4.2.4

Rewrite the expression.

$- 1 (\frac{2730}{2})$

Step 4.4.3

Simplify the expression.

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Step 4.4.3.1

Divide $2730$ by $2$ .

$- 1 \cdot 1365$

Step 4.4.3.2

Multiply $- 1$ by $1365$ .

$- 1365$

Step 5

Add the results of the summations.

$720 - 1365$

Step 6

Subtract $1365$ from $720$ .

$- 645$