Calculus Examples

$\sum_{n = 1}^{15} \frac{1}{2} n + 4$

Step 1

Simplify the summation.

Tap for more steps...

Step 1.1

Combine $\frac{1}{2}$ and $n$ .

$\frac{n}{2} + 4$

Step 1.2

Rewrite the summation.

$\sum_{n = 1}^{15} \frac{n}{2} + 4$

Step 2

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

$\sum_{n = 1}^{15} \frac{n}{2} + 4 = \sum_{n = 1}^{15} \frac{n}{2} + \sum_{n = 1}^{15} 4$

Step 3

Evaluate $\sum_{n = 1}^{15} \frac{n}{2}$ .

Tap for more steps...

Step 3.1

Factor $\frac{1}{2}$ out of the summation.

$(\frac{1}{2}) \sum_{n = 1}^{15} n$

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Simplify.

Tap for more steps...

Step 3.4.1

Simplify the expression.

Tap for more steps...

Step 3.4.1.1

Add $15$ and $1$ .

$\frac{1}{2} \cdot \frac{15 \cdot 16}{2}$

Step 3.4.1.2

Multiply $15$ by $16$ .

$\frac{1}{2} \cdot \frac{240}{2}$

Step 3.4.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.4.2.1

Factor $2$ out of $240$ .

$\frac{1}{2} \cdot \frac{2 (120)}{2}$

Step 3.4.2.2

Cancel the common factor.

$\frac{1}{2} \cdot \frac{2 \cdot 120}{2}$

Step 3.4.2.3

Rewrite the expression.

$\frac{120}{2}$

Step 3.4.3

Divide $120$ by $2$ .

$60$

Step 4

Add the results of the summations.

$60 + 60$

Step 5

Add $60$ and $60$ .

$120$