Calculus Examples

$\sum_{n = 2}^{10} - 2 n - 3$

Step 1

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

$\sum_{n = 2}^{10} - 2 n - 3 = \sum_{n = 1}^{10} - 2 n - 3 - \sum_{n = 1}^{1} - 2 n - 3$

Step 2

Evaluate $\sum_{n = 1}^{10} - 2 n - 3$ .

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Step 2.1

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

$\sum_{n = 1}^{10} - 2 n - 3 = - 2 \sum_{n = 1}^{10} n + \sum_{n = 1}^{10} - 3$

Step 2.2

Evaluate $2 \sum_{n = 1}^{10} n$ .

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Simplify.

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Step 2.2.3.1

Simplify the expression.

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Step 2.2.3.1.1

Add $10$ and $1$ .

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

Step 2.2.3.1.2

Multiply $10$ by $11$ .

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

Step 2.2.3.2

Cancel the common factor of $2$ .

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Step 2.2.3.2.1

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{110}{2}$

Step 2.2.3.2.2

Cancel the common factor.

$2 \cdot - 1 \frac{110}{2}$

Step 2.2.3.2.3

Rewrite the expression.

$- 1 \cdot 110$

Step 2.2.3.3

Multiply $- 1$ by $110$ .

$- 110$

Step 2.3

Evaluate $\sum_{n = 1}^{10} - 3$ .

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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.

$(- 3) (10)$

Step 2.3.3

Multiply $- 3$ by $10$ .

$- 30$

Step 2.4

Add the results of the summations.

$- 110 - 30$

Step 2.5

Subtract $30$ from $- 110$ .

$- 140$

Step 3

Evaluate $\sum_{n = 1}^{1} - 2 n - 3$ .

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Step 3.1

Expand the series for each value of $n$ .

$- 2 \cdot 1 - 3$

Step 3.2

Simplify.

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Step 3.2.1

Multiply $- 2$ by $1$ .

$- 2 - 3$

Step 3.2.2

Subtract $3$ from $- 2$ .

$- 5$

Step 4

Replace the summations with the values found.

$- 140 + 5$

Step 5

Add $- 140$ and $5$ .

$- 135$