Precalculus Examples

$\sum_{1}^{10} {(x - 14)}^{2}$

Step 1

Simplify the summation.

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Step 1.1

Rewrite ${(x - 14)}^{2}$ as $(x - 14) (x - 14)$ .

$(x - 14) (x - 14)$

Step 1.2

Expand $(x - 14) (x - 14)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$x (x - 14) - 14 (x - 14)$

Step 1.2.2

Apply the distributive property.

$x \cdot x + x \cdot - 14 - 14 (x - 14)$

Step 1.2.3

Apply the distributive property.

$x \cdot x + x \cdot - 14 - 14 x - 14 \cdot - 14$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 14 - 14 x - 14 \cdot - 14$

Step 1.3.1.2

Move $- 14$ to the left of $x$ .

$x^{2} - 14 \cdot x - 14 x - 14 \cdot - 14$

Step 1.3.1.3

Multiply $- 14$ by $- 14$ .

$x^{2} - 14 x - 14 x + 196$

Step 1.3.2

Subtract $14 x$ from $- 14 x$ .

$x^{2} - 28 x + 196$

Step 1.4

Rewrite the summation.

$\sum_{x = 1}^{10} x^{2} - 28 x + 196$

Step 2

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

$\sum_{x = 1}^{10} x^{2} - 28 x + 196 = \sum_{x = 1}^{10} x^{2} - 28 \sum_{x = 1}^{10} x + \sum_{x = 1}^{10} 196$

Step 3

Evaluate $\sum_{x = 1}^{10} x^{2}$ .

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Step 3.1

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 3.2

Substitute the values into the formula.

$\frac{10 (10 + 1) (2 \cdot 10 + 1)}{6}$

Step 3.3

Simplify.

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Step 3.3.1

Cancel the common factor of $10$ and $6$ .

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Step 3.3.1.1

Factor $2$ out of $10 (10 + 1) (2 \cdot 10 + 1)$ .

$\frac{2 (5 (10 + 1) (2 \cdot 10 + 1))}{6}$

Step 3.3.1.2

Cancel the common factors.

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Step 3.3.1.2.1

Factor $2$ out of $6$ .

$\frac{2 (5 (10 + 1) (2 \cdot 10 + 1))}{2 (3)}$

Step 3.3.1.2.2

Cancel the common factor.

$\frac{2 (5 (10 + 1) (2 \cdot 10 + 1))}{2 \cdot 3}$

Step 3.3.1.2.3

Rewrite the expression.

$\frac{5 (10 + 1) (2 \cdot 10 + 1)}{3}$

Step 3.3.2

Simplify the numerator.

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Step 3.3.2.1

Multiply $2$ by $10$ .

$\frac{5 (10 + 1) (20 + 1)}{3}$

Step 3.3.2.2

Add $10$ and $1$ .

$\frac{5 \cdot 11 (20 + 1)}{3}$

Step 3.3.2.3

Multiply $5$ by $11$ .

$\frac{55 (20 + 1)}{3}$

Step 3.3.2.4

Add $20$ and $1$ .

$\frac{55 \cdot 21}{3}$

Step 3.3.3

Simplify the expression.

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Step 3.3.3.1

Multiply $55$ by $21$ .

$\frac{1155}{3}$

Step 3.3.3.2

Divide $1155$ by $3$ .

$385$

Step 4

Evaluate $28 \sum_{x = 1}^{10} x$ .

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Step 4.1

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

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

Step 4.2

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

$(- 28) (\frac{10 (10 + 1)}{2})$

Step 4.3

Simplify.

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Step 4.3.1

Simplify the expression.

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Step 4.3.1.1

Add $10$ and $1$ .

$- 28 \frac{10 \cdot 11}{2}$

Step 4.3.1.2

Multiply $10$ by $11$ .

$- 28 (\frac{110}{2})$

Step 4.3.2

Cancel the common factor of $2$ .

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Step 4.3.2.1

Factor $2$ out of $- 28$ .

$2 (- 14) \frac{110}{2}$

Step 4.3.2.2

Cancel the common factor.

$2 \cdot - 14 \frac{110}{2}$

Step 4.3.2.3

Rewrite the expression.

$- 14 \cdot 110$

Step 4.3.3

Multiply $- 14$ by $110$ .

$- 1540$

Step 5

Evaluate $\sum_{x = 1}^{10} 196$ .

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Step 5.1

The formula for the summation of a constant is:

$\sum_{k = 1}^{n} c = c n$

Step 5.2

Substitute the values into the formula.

$(196) (10)$

Step 5.3

Multiply $196$ by $10$ .

$1960$

Step 6

Add the results of the summations.

$385 - 1540 + 1960$

Step 7

Simplify.

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Step 7.1

Subtract $1540$ from $385$ .

$- 1155 + 1960$

Step 7.2

Add $- 1155$ and $1960$ .

$805$