Calculus Examples

$\sum_{i = 1}^{17} {(i - 1)}^{2}$

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.

$\sum_{i = 1}^{17} i^{2} - 2 i + 1$

Step 2

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

$\sum_{i = 1}^{17} i^{2} - 2 i + 1 = \sum_{i = 1}^{17} i^{2} - 2 \sum_{i = 1}^{17} i + \sum_{i = 1}^{17} 1$

Step 3

Evaluate $\sum_{i = 1}^{17} i^{2}$ .

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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.

$\frac{17 (17 + 1) (2 \cdot 17 + 1)}{6}$

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 $17$ and $1$ .

$\frac{17 \cdot 18 (2 \cdot 17 + 1)}{6}$

Step 3.3.1.2

Combine exponents.

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Step 3.3.1.2.1

Multiply $17$ by $18$ .

$\frac{306 (2 \cdot 17 + 1)}{6}$

Step 3.3.1.2.2

Multiply $2$ by $17$ .

$\frac{306 (34 + 1)}{6}$

Step 3.3.1.3

Add $34$ and $1$ .

$\frac{306 \cdot 35}{6}$

Step 3.3.2

Simplify the expression.

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Step 3.3.2.1

Multiply $306$ by $35$ .

$\frac{10710}{6}$

Step 3.3.2.2

Divide $10710$ by $6$ .

$1785$

Step 4

Evaluate $2 \sum_{i = 1}^{17} i$ .

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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.

$(- 2) (\frac{17 (17 + 1)}{2})$

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 $17$ and $1$ .

$- 2 \frac{17 \cdot 18}{2}$

Step 4.3.1.2

Multiply $17$ by $18$ .

$- 2 (\frac{306}{2})$

Step 4.3.2

Cancel the common factor of $2$ .

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Step 4.3.2.1

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{306}{2}$

Step 4.3.2.2

Cancel the common factor.

$2 \cdot - 1 \frac{306}{2}$

Step 4.3.2.3

Rewrite the expression.

$- 1 \cdot 306$

Step 4.3.3

Multiply $- 1$ by $306$ .

$- 306$

Step 5

Evaluate $\sum_{i = 1}^{17} 1$ .

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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.

$(1) (17)$

Step 5.3

Multiply $17$ by $1$ .

$17$

Step 6

Add the results of the summations.

$1785 - 306 + 17$

Step 7

Simplify.

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Step 7.1

Subtract $306$ from $1785$ .

$1479 + 17$

Step 7.2

Add $1479$ and $17$ .

$1496$