Calculus Examples

$\sum_{k = 5}^{10} - 5 k$

Step 1

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

$\sum_{k = 5}^{10} - 5 k = \sum_{k = 1}^{10} - 5 k - \sum_{k = 1}^{4} - 5 k$

Step 2

Evaluate $\sum_{k = 1}^{10} - 5 k$ .

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Step 2.1

Factor $- 5$ out of the summation.

$(- 5) \sum_{k = 1}^{10} k$

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Simplify.

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Step 2.4.1

Add $10$ and $1$ .

$- 5 \frac{10 \cdot 11}{2}$

Step 2.4.2

Multiply $10$ by $11$ .

$- 5 (\frac{110}{2})$

Step 2.4.3

Divide $110$ by $2$ .

$- 5 \cdot 55$

Step 2.4.4

Multiply $- 5$ by $55$ .

$- 275$

Step 3

Evaluate $\sum_{k = 1}^{4} - 5 k$ .

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Step 3.1

Expand the series for each value of $k$ .

$- 5 \cdot 1 - 5 \cdot 2 - 5 \cdot 3 - 5 \cdot 4$

Step 3.2

Simplify.

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Step 3.2.1

Multiply $- 5$ by $1$ .

$- 5 - 5 \cdot 2 - 5 \cdot 3 - 5 \cdot 4$

Step 3.2.2

Multiply $- 5$ by $2$ .

$- 5 - 10 - 5 \cdot 3 - 5 \cdot 4$

Step 3.2.3

Subtract $10$ from $- 5$ .

$- 15 - 5 \cdot 3 - 5 \cdot 4$

Step 3.2.4

Multiply $- 5$ by $3$ .

$- 15 - 15 - 5 \cdot 4$

Step 3.2.5

Subtract $15$ from $- 15$ .

$- 30 - 5 \cdot 4$

Step 3.2.6

Multiply $- 5$ by $4$ .

$- 30 - 20$

Step 3.2.7

Subtract $20$ from $- 30$ .

$- 50$

Step 4

Replace the summations with the values found.

$- 275 + 50$

Step 5

Add $- 275$ and $50$ .

$- 225$