Precalculus Examples

$\sum_{n = 11}^{30} n$

Step 1

Split the summation to make the starting value of $n$ equal to $1$ .

$\sum_{n = 11}^{30} n = \sum_{n = 1}^{30} n - \sum_{n = 1}^{10} n$

Step 2

Evaluate $\sum_{n = 1}^{30} n$ .

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Step 2.1

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

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

Step 2.2

Substitute the values into the formula.

$\frac{30 (30 + 1)}{2}$

Step 2.3

Simplify.

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Step 2.3.1

Cancel the common factor of $30$ and $2$ .

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Step 2.3.1.1

Factor $2$ out of $30 (30 + 1)$ .

$\frac{2 (15 (30 + 1))}{2}$

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

Factor $2$ out of $2$ .

$\frac{2 (15 (30 + 1))}{2 (1)}$

Step 2.3.1.2.2

Cancel the common factor.

$\frac{2 (15 (30 + 1))}{2 \cdot 1}$

Step 2.3.1.2.3

Rewrite the expression.

$\frac{15 (30 + 1)}{1}$

Step 2.3.1.2.4

Divide $15 (30 + 1)$ by $1$ .

$15 (30 + 1)$

Step 2.3.2

Simplify the expression.

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Step 2.3.2.1

Add $30$ and $1$ .

$15 \cdot 31$

Step 2.3.2.2

Multiply $15$ by $31$ .

$465$

Step 3

Evaluate $\sum_{n = 1}^{10} n$ .

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Step 3.1

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

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

Step 3.2

Substitute the values into the formula.

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

Step 3.3

Simplify.

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Step 3.3.1

Cancel the common factor of $10$ and $2$ .

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Step 3.3.1.1

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

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

Step 3.3.1.2

Cancel the common factors.

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Step 3.3.1.2.1

Factor $2$ out of $2$ .

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

Step 3.3.1.2.2

Cancel the common factor.

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

Step 3.3.1.2.3

Rewrite the expression.

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

Step 3.3.1.2.4

Divide $5 (10 + 1)$ by $1$ .

$5 (10 + 1)$

Step 3.3.2

Simplify the expression.

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Step 3.3.2.1

Add $10$ and $1$ .

$5 \cdot 11$

Step 3.3.2.2

Multiply $5$ by $11$ .

$55$

Step 4

Replace the summations with the values found.

$465 - 55$

Step 5

Subtract $55$ from $465$ .

$410$