Calculus Examples

$\sum_{n = 1}^{12} 2 n + 1$

Step 1

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

$\sum_{n = 1}^{12} 2 n + 1 = 2 \sum_{n = 1}^{12} n + \sum_{n = 1}^{12} 1$

Step 2

Evaluate $2 \sum_{n = 1}^{12} n$ .

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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 and make sure to multiply by the front term.

$(2) (\frac{12 (12 + 1)}{2})$

Step 2.3

Simplify.

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Step 2.3.1

Simplify the expression.

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Step 2.3.1.1

Add $12$ and $1$ .

$2 \frac{12 \cdot 13}{2}$

Step 2.3.1.2

Multiply $12$ by $13$ .

$2 (\frac{156}{2})$

Step 2.3.2

Cancel the common factor of $2$ .

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Step 2.3.2.1

Cancel the common factor.

$2 (\frac{156}{2})$

Step 2.3.2.2

Rewrite the expression.

$156$

Step 3

Evaluate $\sum_{n = 1}^{12} 1$ .

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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.

$(1) (12)$

Step 3.3

Multiply $12$ by $1$ .

$12$

Step 4

Add the results of the summations.

$156 + 12$

Step 5

Add $156$ and $12$ .

$168$