Precalculus Examples

$\sum_{n = 1}^{500} n + 8$

Step 1

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

$\sum_{n = 1}^{500} n + 8 = \sum_{n = 1}^{500} n + \sum_{n = 1}^{500} 8$

Step 2

Evaluate $\sum_{n = 1}^{500} n$ .

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Step 2.1

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

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

Step 2.2

Substitute the values into the formula.

$\frac{500 (500 + 1)}{2}$

Step 2.3

Simplify.

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Step 2.3.1

Cancel the common factor of $500$ and $2$ .

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Step 2.3.1.1

Factor $2$ out of $500 (500 + 1)$ .

$\frac{2 (250 (500 + 1))}{2}$

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

Factor $2$ out of $2$ .

$\frac{2 (250 (500 + 1))}{2 (1)}$

Step 2.3.1.2.2

Cancel the common factor.

$\frac{2 (250 (500 + 1))}{2 \cdot 1}$

Step 2.3.1.2.3

Rewrite the expression.

$\frac{250 (500 + 1)}{1}$

Step 2.3.1.2.4

Divide $250 (500 + 1)$ by $1$ .

$250 (500 + 1)$

Step 2.3.2

Simplify the expression.

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Step 2.3.2.1

Add $500$ and $1$ .

$250 \cdot 501$

Step 2.3.2.2

Multiply $250$ by $501$ .

$125250$

Step 3

Evaluate $\sum_{n = 1}^{500} 8$ .

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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.

$(8) (500)$

Step 3.3

Multiply $8$ by $500$ .

$4000$

Step 4

Add the results of the summations.

$125250 + 4000$

Step 5

Add $125250$ and $4000$ .

$129250$