Precalculus Examples

$\sum_{k = 1}^{3} k (k + 5)$

Step 1

Simplify the summation.

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Step 1.1

Apply the distributive property.

$k \cdot k + k \cdot 5$

Step 1.2

Multiply $k$ by $k$ .

$k^{2} + k \cdot 5$

Step 1.3

Move $5$ to the left of $k$ .

$k^{2} + 5 k$

Step 1.4

Rewrite the summation.

$\sum_{k = 1}^{3} k^{2} + 5 k$

Step 2

The formula for the summation of a polynomial with degree $2$ is:

$\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

Step 3

Substitute the values into the formula.

$\frac{3 (3 + 1) (2 \cdot 3 + 1)}{6}$

Step 4

Simplify.

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Step 4.1

Cancel the common factor of $3$ and $6$ .

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Step 4.1.1

Factor $3$ out of $3 (3 + 1) (2 \cdot 3 + 1)$ .

$\frac{3 ((3 + 1) (2 \cdot 3 + 1))}{6}$

Step 4.1.2

Cancel the common factors.

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Step 4.1.2.1

Factor $3$ out of $6$ .

$\frac{3 ((3 + 1) (2 \cdot 3 + 1))}{3 \cdot 2}$

Step 4.1.2.2

Cancel the common factor.

$\frac{3 ((3 + 1) (2 \cdot 3 + 1))}{3 \cdot 2}$

Step 4.1.2.3

Rewrite the expression.

$\frac{(3 + 1) (2 \cdot 3 + 1)}{2}$

Step 4.2

Simplify the numerator.

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Step 4.2.1

Add $3$ and $1$ .

$\frac{4 (2 \cdot 3 + 1)}{2}$

Step 4.2.2

Multiply $2$ by $3$ .

$\frac{4 (6 + 1)}{2}$

Step 4.2.3

Add $6$ and $1$ .

$\frac{4 \cdot 7}{2}$

Step 4.3

Simplify the expression.

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Step 4.3.1

Multiply $4$ by $7$ .

$\frac{28}{2}$

Step 4.3.2

Divide $28$ by $2$ .

$14$