Calculus Examples

$\sum_{k = 1}^{20} 100 (k^{2} - 5 k + 1)$

Step 1

Simplify the summation.

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Step 1.1

Apply the distributive property.

$100 k^{2} + 100 (- 5 k) + 100 \cdot 1$

Step 1.2

Simplify.

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Step 1.2.1

Multiply $- 5$ by $100$ .

$100 k^{2} - 500 k + 100 \cdot 1$

Step 1.2.2

Multiply $100$ by $1$ .

$100 k^{2} - 500 k + 100$

Step 1.3

Rewrite the summation.

$\sum_{k = 1}^{20} 100 k^{2} - 500 k + 100$

Step 2

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

$\sum_{k = 1}^{20} 100 k^{2} - 500 k + 100 = 100 \sum_{k = 1}^{20} k^{2} - 500 \sum_{k = 1}^{20} k + \sum_{k = 1}^{20} 100$

Step 3

Evaluate $100 \sum_{k = 1}^{20} k^{2}$ .

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Step 3.1

The formula for the summation of a polynomial with degree $2$ is:

$\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

Step 3.2

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

$(100) (\frac{20 (20 + 1) (2 \cdot 20 + 1)}{6})$

Step 3.3

Simplify.

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Step 3.3.1

Simplify the numerator.

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Step 3.3.1.1

Add $20$ and $1$ .

$100 \frac{20 \cdot 21 (2 \cdot 20 + 1)}{6}$

Step 3.3.1.2

Combine exponents.

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Step 3.3.1.2.1

Multiply $20$ by $21$ .

$100 \frac{420 (2 \cdot 20 + 1)}{6}$

Step 3.3.1.2.2

Multiply $2$ by $20$ .

$100 \frac{420 (40 + 1)}{6}$

Step 3.3.1.3

Add $40$ and $1$ .

$100 \frac{420 \cdot 41}{6}$

Step 3.3.2

Simplify terms.

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Step 3.3.2.1

Multiply $420$ by $41$ .

$100 (\frac{17220}{6})$

Step 3.3.2.2

Cancel the common factor of $2$ .

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Step 3.3.2.2.1

Factor $2$ out of $100$ .

$2 (50) \frac{17220}{6}$

Step 3.3.2.2.2

Factor $2$ out of $6$ .

$2 \cdot 50 \frac{17220}{2 \cdot 3}$

Step 3.3.2.2.3

Cancel the common factor.

$2 \cdot 50 \frac{17220}{2 \cdot 3}$

Step 3.3.2.2.4

Rewrite the expression.

$50 (\frac{17220}{3})$

Step 3.3.2.3

Combine $50$ and $\frac{17220}{3}$ .

$\frac{50 \cdot 17220}{3}$

Step 3.3.2.4

Simplify the expression.

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Step 3.3.2.4.1

Multiply $50$ by $17220$ .

$\frac{861000}{3}$

Step 3.3.2.4.2

Divide $861000$ by $3$ .

$287000$

Step 4

Evaluate $500 \sum_{k = 1}^{20} k$ .

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Step 4.1

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

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

Step 4.2

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

$(- 500) (\frac{20 (20 + 1)}{2})$

Step 4.3

Simplify.

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Step 4.3.1

Simplify the expression.

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Step 4.3.1.1

Add $20$ and $1$ .

$- 500 \frac{20 \cdot 21}{2}$

Step 4.3.1.2

Multiply $20$ by $21$ .

$- 500 (\frac{420}{2})$

Step 4.3.2

Cancel the common factor of $2$ .

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Step 4.3.2.1

Factor $2$ out of $- 500$ .

$2 (- 250) \frac{420}{2}$

Step 4.3.2.2

Cancel the common factor.

$2 \cdot - 250 \frac{420}{2}$

Step 4.3.2.3

Rewrite the expression.

$- 250 \cdot 420$

Step 4.3.3

Multiply $- 250$ by $420$ .

$- 105000$

Step 5

Evaluate $\sum_{k = 1}^{20} 100$ .

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Step 5.1

The formula for the summation of a constant is:

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

Step 5.2

Substitute the values into the formula.

$(100) (20)$

Step 5.3

Multiply $100$ by $20$ .

$2000$

Step 6

Add the results of the summations.

$287000 - 105000 + 2000$

Step 7

Simplify.

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Step 7.1

Subtract $105000$ from $287000$ .

$182000 + 2000$

Step 7.2

Add $182000$ and $2000$ .

$184000$