Precalculus Examples

$\sum_{n = 1}^{20} \frac{5}{2} n - 3$

Step 1

Simplify the summation.

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Step 1.1

Combine $\frac{5}{2}$ and $n$ .

$\frac{5 n}{2} - 3$

Step 1.2

Rewrite the summation.

$\sum_{n = 1}^{20} \frac{5 n}{2} - 3$

Step 2

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

$\sum_{n = 1}^{20} \frac{5 n}{2} - 3 = \sum_{n = 1}^{20} \frac{5 n}{2} + \sum_{n = 1}^{20} - 3$

Step 3

Evaluate $\sum_{n = 1}^{20} \frac{5 n}{2}$ .

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Step 3.1

Factor $\frac{5}{2}$ out of the summation.

$(\frac{5}{2}) \sum_{n = 1}^{20} n$

Step 3.2

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

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

Step 3.3

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

$(\frac{5}{2}) (\frac{20 (20 + 1)}{2})$

Step 3.4

Simplify.

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Step 3.4.1

Simplify the expression.

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Step 3.4.1.1

Add $20$ and $1$ .

$\frac{5}{2} \cdot \frac{20 \cdot 21}{2}$

Step 3.4.1.2

Multiply $20$ by $21$ .

$\frac{5}{2} \cdot \frac{420}{2}$

Step 3.4.2

Cancel the common factor of $2$ .

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Step 3.4.2.1

Factor $2$ out of $420$ .

$\frac{5}{2} \cdot \frac{2 (210)}{2}$

Step 3.4.2.2

Cancel the common factor.

$\frac{5}{2} \cdot \frac{2 \cdot 210}{2}$

Step 3.4.2.3

Rewrite the expression.

$5 (\frac{210}{2})$

Step 3.4.3

Combine $5$ and $\frac{210}{2}$ .

$\frac{5 \cdot 210}{2}$

Step 3.4.4

Simplify the expression.

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Step 3.4.4.1

Multiply $5$ by $210$ .

$\frac{1050}{2}$

Step 3.4.4.2

Divide $1050$ by $2$ .

$525$

Step 4

Evaluate $\sum_{n = 1}^{20} - 3$ .

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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.

$(- 3) (20)$

Step 4.3

Multiply $- 3$ by $20$ .

$- 60$

Step 5

Add the results of the summations.

$525 - 60$

Step 6

Subtract $60$ from $525$ .

$465$