Calculus Examples

$\sum_{i = 1}^{20} {(i - 1)}^{2}$

Step 1

Simplify the summation.

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Step 1.1

Rewrite

${(i - 1)}^{2}$ as

$(i - 1) (i - 1)$ .

$(i - 1) (i - 1)$

Step 1.2

Expand

$(i - 1) (i - 1)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$i (i - 1) - 1 (i - 1)$

Step 1.2.2

Apply the distributive property.

$i \cdot i + i \cdot - 1 - 1 (i - 1)$

Step 1.2.3

Apply the distributive property.

$i \cdot i + i \cdot - 1 - 1 i - 1 \cdot - 1$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply

$i$ by

$i$ .

$i^{2} + i \cdot - 1 - 1 i - 1 \cdot - 1$

Step 1.3.1.2

Move

$- 1$ to the left of

$i$ .

$i^{2} - 1 \cdot i - 1 i - 1 \cdot - 1$

Step 1.3.1.3

Rewrite

$- 1 i$ as

$- i$ .

$i^{2} - i - 1 i - 1 \cdot - 1$

Step 1.3.1.4

Rewrite

$- 1 i$ as

$- i$ .

$i^{2} - i - i - 1 \cdot - 1$

Step 1.3.1.5

Multiply

$- 1$ by

$- 1$ .

$i^{2} - i - i + 1$

Step 1.3.2

Subtract

$i$ from

$- i$ .

$i^{2} - 2 i + 1$

Step 1.4

Rewrite the summation.

$\sum_{i = 1}^{20} i^{2} - 2 i + 1$

Step 2

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

$\sum_{i = 1}^{20} i^{2} - 2 i + 1 = \sum_{i = 1}^{20} i^{2} - 2 \sum_{i = 1}^{20} i + \sum_{i = 1}^{20} 1$

Step 3

Evaluate

$\sum_{i = 1}^{20} i^{2}$ .

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Step 3.1

The formula for the summation of a polynomial with degree

$2$ is:

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

Step 3.2

Substitute the values into the formula.

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

Step 3.3

Simplify.

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Step 3.3.1

Cancel the common factor of

$20$ and

$6$ .

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Step 3.3.1.1

Factor

$2$ out of

$20 (20 + 1) (2 \cdot 20 + 1)$ .

$\frac{2 (10 (20 + 1) (2 \cdot 20 + 1))}{6}$

Step 3.3.1.2

Cancel the common factors.

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Step 3.3.1.2.1

Factor

$2$ out of

$6$ .

$\frac{2 (10 (20 + 1) (2 \cdot 20 + 1))}{2 (3)}$

Step 3.3.1.2.2

Cancel the common factor.

$\frac{2 (10 (20 + 1) (2 \cdot 20 + 1))}{2 \cdot 3}$

Step 3.3.1.2.3

Rewrite the expression.

$\frac{10 (20 + 1) (2 \cdot 20 + 1)}{3}$

Step 3.3.2

Simplify the numerator.

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Step 3.3.2.1

Multiply

$2$ by

$20$ .

$\frac{10 (20 + 1) (40 + 1)}{3}$

Step 3.3.2.2

Add

$20$ and

$1$ .

$\frac{10 \cdot 21 (40 + 1)}{3}$

Step 3.3.2.3

Multiply

$10$ by

$21$ .

$\frac{210 (40 + 1)}{3}$

Step 3.3.2.4

Add

$40$ and

$1$ .

$\frac{210 \cdot 41}{3}$

Step 3.3.3

Simplify the expression.

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Step 3.3.3.1

Multiply

$210$ by

$41$ .

$\frac{8610}{3}$

Step 3.3.3.2

Divide

$8610$ by

$3$ .

$2870$

Step 4

Evaluate

$2 \sum_{i = 1}^{20} i$ .

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Step 4.1

The formula for the summation of a polynomial with degree

$1$ is:

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

Step 4.2

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

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

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

$1$ .

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

Step 4.3.1.2

Multiply

$20$ by

$21$ .

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

Step 4.3.2

Cancel the common factor of

$2$ .

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Step 4.3.2.1

Factor

$2$ out of

$- 2$ .

$2 (- 1) \frac{420}{2}$

Step 4.3.2.2

Cancel the common factor.

$2 \cdot - 1 \frac{420}{2}$

Step 4.3.2.3

Rewrite the expression.

$- 1 \cdot 420$

Step 4.3.3

Multiply

$- 1$ by

$420$ .

$- 420$

Step 5

Evaluate

$\sum_{i = 1}^{20} 1$ .

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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.

$(1) (20)$

Step 5.3

Multiply

$20$ by

$1$ .

$20$

Step 6

Add the results of the summations.

$2870 - 420 + 20$

Step 7

Simplify.

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Step 7.1

Subtract

$420$ from

$2870$ .

$2450 + 20$

Step 7.2

Add

$2450$ and

$20$ .

$2470$