Precalculus Examples

$\sum_{n = 1}^{31} 300 + {(n - 1)}^{2}$

Step 1

Simplify the summation.

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Step 1.1

Simplify each term.

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Step 1.1.1

Rewrite ${(n - 1)}^{2}$ as $(n - 1) (n - 1)$ .

$300 + (n - 1) (n - 1)$

Step 1.1.2

Expand $(n - 1) (n - 1)$ using the FOIL Method.

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Step 1.1.2.1

Apply the distributive property.

$300 + n (n - 1) - 1 (n - 1)$

Step 1.1.2.2

Apply the distributive property.

$300 + n \cdot n + n \cdot - 1 - 1 (n - 1)$

Step 1.1.2.3

Apply the distributive property.

$300 + n \cdot n + n \cdot - 1 - 1 n - 1 \cdot - 1$

Step 1.1.3

Simplify and combine like terms.

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Step 1.1.3.1

Simplify each term.

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Step 1.1.3.1.1

Multiply $n$ by $n$ .

$300 + n^{2} + n \cdot - 1 - 1 n - 1 \cdot - 1$

Step 1.1.3.1.2

Move $- 1$ to the left of $n$ .

$300 + n^{2} - 1 \cdot n - 1 n - 1 \cdot - 1$

Step 1.1.3.1.3

Rewrite $- 1 n$ as $- n$ .

$300 + n^{2} - n - 1 n - 1 \cdot - 1$

Step 1.1.3.1.4

Rewrite $- 1 n$ as $- n$ .

$300 + n^{2} - n - n - 1 \cdot - 1$

Step 1.1.3.1.5

Multiply $- 1$ by $- 1$ .

$300 + n^{2} - n - n + 1$

Step 1.1.3.2

Subtract $n$ from $- n$ .

$300 + n^{2} - 2 n + 1$

Step 1.2

Add $300$ and $1$ .

$n^{2} - 2 n + 301$

Step 1.3

Rewrite the summation.

$\sum_{n = 1}^{31} n^{2} - 2 n + 301$

Step 2

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

$\sum_{n = 1}^{31} n^{2} - 2 n + 301 = \sum_{n = 1}^{31} n^{2} - 2 \sum_{n = 1}^{31} n + \sum_{n = 1}^{31} 301$

Step 3

Evaluate $\sum_{n = 1}^{31} n^{2}$ .

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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.

$\frac{31 (31 + 1) (2 \cdot 31 + 1)}{6}$

Step 3.3

Simplify.

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Step 3.3.1

Simplify the numerator.

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Step 3.3.1.1

Add $31$ and $1$ .

$\frac{31 \cdot 32 (2 \cdot 31 + 1)}{6}$

Step 3.3.1.2

Combine exponents.

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Step 3.3.1.2.1

Multiply $31$ by $32$ .

$\frac{992 (2 \cdot 31 + 1)}{6}$

Step 3.3.1.2.2

Multiply $2$ by $31$ .

$\frac{992 (62 + 1)}{6}$

Step 3.3.1.3

Add $62$ and $1$ .

$\frac{992 \cdot 63}{6}$

Step 3.3.2

Simplify the expression.

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Step 3.3.2.1

Multiply $992$ by $63$ .

$\frac{62496}{6}$

Step 3.3.2.2

Divide $62496$ by $6$ .

$10416$

Step 4

Evaluate $2 \sum_{n = 1}^{31} n$ .

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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.

$(- 2) (\frac{31 (31 + 1)}{2})$

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 $31$ and $1$ .

$- 2 \frac{31 \cdot 32}{2}$

Step 4.3.1.2

Multiply $31$ by $32$ .

$- 2 (\frac{992}{2})$

Step 4.3.2

Cancel the common factor of $2$ .

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Step 4.3.2.1

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{992}{2}$

Step 4.3.2.2

Cancel the common factor.

$2 \cdot - 1 \frac{992}{2}$

Step 4.3.2.3

Rewrite the expression.

$- 1 \cdot 992$

Step 4.3.3

Multiply $- 1$ by $992$ .

$- 992$

Step 5

Evaluate $\sum_{n = 1}^{31} 301$ .

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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.

$(301) (31)$

Step 5.3

Multiply $301$ by $31$ .

$9331$

Step 6

Add the results of the summations.

$10416 - 992 + 9331$

Step 7

Simplify.

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Step 7.1

Subtract $992$ from $10416$ .

$9424 + 9331$

Step 7.2

Add $9424$ and $9331$ .

$18755$