Calculus Examples

$\sum_{3}^{8} 15 - 2 n$

Step 1

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

$\sum_{3}^{8} 15 - 2 n = \sum_{n = 1}^{8} 15 - 2 n - \sum_{n = 1}^{2} 15 - 2 n$

Step 2

Evaluate $\sum_{n = 1}^{8} 15 - 2 n$ .

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

$\sum_{n = 1}^{8} 15 - 2 n = \sum_{n = 1}^{8} 15 - 2 \sum_{n = 1}^{8} n$

Evaluate $\sum_{n = 1}^{8} 15$ .

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The formula for the summation of a constant is:

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

Substitute the values into the formula.

$(15) (8)$

Multiply $15$ by $8$ .

$120$

Evaluate $2 \sum_{n = 1}^{8} n$ .

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

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

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

$(- 2) (\frac{8 (8 + 1)}{2})$

Simplify.

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Simplify the expression.

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Add $8$ and $1$ .

$- 2 \frac{8 \cdot 9}{2}$

Multiply $8$ by $9$ .

$- 2 (\frac{72}{2})$

Cancel the common factor of $2$ .

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Factor $2$ out of $- 2$ .

$2 (- 1) \frac{72}{2}$

Cancel the common factor.

$2 \cdot - 1 \frac{72}{2}$

Rewrite the expression.

$- 1 \cdot 72$

Multiply $- 1$ by $72$ .

$- 72$

Add the results of the summations.

$120 - 72$

Subtract $72$ from $120$ .

$48$

Step 3

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

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Expand the series for each value of $n$ .

$15 - 2 \cdot 1 + 15 - 2 \cdot 2$

Simplify.

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Multiply $- 2$ by $1$ .

$15 - 2 + 15 - 2 \cdot 2$

Subtract $2$ from $15$ .

$13 + 15 - 2 \cdot 2$

Multiply $- 2$ by $2$ .

$13 + 15 - 4$

Subtract $4$ from $15$ .

$13 + 11$

Add $13$ and $11$ .

$24$

Step 4

Replace the summations with the values found.

$48 - 24$

Step 5

Subtract $24$ from $48$ .

$24$