Precalculus Examples

$\sum_{i = 1}^{20} 3 - 2 (i - 3)$

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.

$3 - 2 i - 2 \cdot - 3$

Step 1.1.2

Multiply $- 2$ by $- 3$ .

$3 - 2 i + 6$

Step 1.2

Add $3$ and $6$ .

$- 2 i + 9$

Step 1.3

Rewrite the summation.

$\sum_{i = 1}^{20} - 2 i + 9$

Step 2

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

$\sum_{i = 1}^{20} - 2 i + 9 = - 2 \sum_{i = 1}^{20} i + \sum_{i = 1}^{20} 9$

Step 3

Evaluate $2 \sum_{i = 1}^{20} i$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Simplify.

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Step 3.3.1

Simplify the expression.

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Step 3.3.1.1

Add $20$ and $1$ .

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

Step 3.3.1.2

Multiply $20$ by $21$ .

$- 2 (\frac{420}{2})$

Step 3.3.2

Cancel the common factor of $2$ .

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Step 3.3.2.1

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{420}{2}$

Step 3.3.2.2

Cancel the common factor.

$2 \cdot - 1 \frac{420}{2}$

Step 3.3.2.3

Rewrite the expression.

$- 1 \cdot 420$

Step 3.3.3

Multiply $- 1$ by $420$ .

$- 420$

Step 4

Evaluate $\sum_{i = 1}^{20} 9$ .

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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.

$(9) (20)$

Step 4.3

Multiply $9$ by $20$ .

$180$

Step 5

Add the results of the summations.

$- 420 + 180$

Step 6

Add $- 420$ and $180$ .

$- 240$