Precalculus Examples

$\sum_{i = 3}^{8} - 5 i + 2$

Step 1

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

$\sum_{k = 3}^{8} - 5 i + 2 = \sum_{k = 1}^{8} - 5 i + 2 - \sum_{k = 1}^{2} - 5 i + 2$

Step 2

Evaluate $\sum_{k = 1}^{8} - 5 i + 2$ .

Tap for more steps...

Step 2.1

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

$\sum_{k = 1}^{8} - 5 i + 2 = - 5 \sum_{k = 1}^{8} i + \sum_{k = 1}^{8} 2$

Step 2.2

Evaluate $5 \sum_{k = 1}^{8} i$ .

Tap for more steps...

Step 2.2.1

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

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

Step 2.2.2

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

$(- 5) (\frac{8 (8 + 1)}{2})$

Step 2.2.3

Simplify.

Tap for more steps...

Step 2.2.3.1

Add $8$ and $1$ .

$- 5 \frac{8 \cdot 9}{2}$

Step 2.2.3.2

Multiply $8$ by $9$ .

$- 5 (\frac{72}{2})$

Step 2.2.3.3

Divide $72$ by $2$ .

$- 5 \cdot 36$

Step 2.2.3.4

Multiply $- 5$ by $36$ .

$- 180$

Step 2.3

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

Tap for more steps...

Step 2.3.1

The formula for the summation of a constant is:

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

Step 2.3.2

Substitute the values into the formula.

$(2) (8)$

Step 2.3.3

Multiply $2$ by $8$ .

$16$

Step 2.4

Add the results of the summations.

$- 180 + 16$

Step 2.5

Add $- 180$ and $16$ .

$- 164$

Step 3

Evaluate $\sum_{k = 1}^{2} - 5 i + 2$ .

Tap for more steps...

Step 3.1

Expand the series for each value of $i$ .

$- 5 \cdot 1 + 2 - 5 \cdot 2 + 2$

Step 3.2

Simplify.

Tap for more steps...

Step 3.2.1

Multiply $- 5$ by $1$ .

$- 5 + 2 - 5 \cdot 2 + 2$

Step 3.2.2

Add $- 5$ and $2$ .

$- 3 - 5 \cdot 2 + 2$

Step 3.2.3

Multiply $- 5$ by $2$ .

$- 3 - 10 + 2$

Step 3.2.4

Add $- 10$ and $2$ .

$- 3 - 8$

Step 3.2.5

Subtract $8$ from $- 3$ .

$- 11$

Step 4

Replace the summations with the values found.

$- 164 + 11$

Step 5

Add $- 164$ and $11$ .

$- 153$