Precalculus Examples

$\sum_{n = 1}^{120} 4 - \frac{1}{3} n$

Step 1

Simplify the summation.

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Step 1.1

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

$4 - \frac{n}{3}$

Step 1.2

Rewrite the summation.

$\sum_{n = 1}^{120} 4 - \frac{n}{3}$

Step 2

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

$\sum_{n = 1}^{120} 4 - \frac{n}{3} = \sum_{n = 1}^{120} 4 + \sum_{n = 1}^{120} - \frac{n}{3}$

Step 3

Evaluate $\sum_{n = 1}^{120} 4$ .

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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.

$(4) (120)$

Step 3.3

Multiply $4$ by $120$ .

$480$

Step 4

Evaluate $\sum_{n = 1}^{120} - \frac{n}{3}$ .

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Step 4.1

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

$(- \frac{1}{3}) \sum_{n = 1}^{120} n$

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.

$(- \frac{1}{3}) (\frac{120 (120 + 1)}{2})$

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 $120$ and $1$ .

$- \frac{1}{3} \cdot \frac{120 \cdot 121}{2}$

Step 4.4.1.2

Multiply $120$ by $121$ .

$- \frac{1}{3} \cdot \frac{14520}{2}$

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.

$\frac{- 1}{3} \cdot \frac{14520}{2}$

Step 4.4.2.2

Factor $3$ out of $14520$ .

$\frac{- 1}{3} \cdot \frac{3 (4840)}{2}$

Step 4.4.2.3

Cancel the common factor.

$\frac{- 1}{3} \cdot \frac{3 \cdot 4840}{2}$

Step 4.4.2.4

Rewrite the expression.

$- 1 (\frac{4840}{2})$

Step 4.4.3

Simplify the expression.

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Step 4.4.3.1

Divide $4840$ by $2$ .

$- 1 \cdot 2420$

Step 4.4.3.2

Multiply $- 1$ by $2420$ .

$- 2420$

Step 5

Add the results of the summations.

$480 - 2420$

Step 6

Subtract $2420$ from $480$ .

$- 1940$