Calculus Examples

$\sum_{i = 1}^{12} 2 i^{2} - 7$

Step 1

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

$\sum_{k = 1}^{12} 2 i^{2} - 7 = 2 \sum_{k = 1}^{12} i^{2} + \sum_{k = 1}^{12} - 7$

Step 2

Evaluate $2 \sum_{k = 1}^{12} i^{2}$ .

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Step 2.1

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 2.2

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

$(2) (\frac{12 (12 + 1) (2 \cdot 12 + 1)}{6})$

Step 2.3

Simplify.

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Step 2.3.1

Simplify the numerator.

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Step 2.3.1.1

Add $12$ and $1$ .

$2 \frac{12 \cdot 13 (2 \cdot 12 + 1)}{6}$

Step 2.3.1.2

Combine exponents.

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Step 2.3.1.2.1

Multiply $12$ by $13$ .

$2 \frac{156 (2 \cdot 12 + 1)}{6}$

Step 2.3.1.2.2

Multiply $2$ by $12$ .

$2 \frac{156 (24 + 1)}{6}$

Step 2.3.1.3

Add $24$ and $1$ .

$2 \frac{156 \cdot 25}{6}$

Step 2.3.2

Reduce the expression by cancelling the common factors.

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Step 2.3.2.1

Multiply $156$ by $25$ .

$2 (\frac{3900}{6})$

Step 2.3.2.2

Cancel the common factor of $2$ .

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Step 2.3.2.2.1

Factor $2$ out of $6$ .

$2 \frac{3900}{2 (3)}$

Step 2.3.2.2.2

Cancel the common factor.

$2 \frac{3900}{2 \cdot 3}$

Step 2.3.2.2.3

Rewrite the expression.

$\frac{3900}{3}$

Step 2.3.2.3

Divide $3900$ by $3$ .

$1300$

Step 3

Evaluate $\sum_{k = 1}^{12} - 7$ .

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Step 3.1

The formula for the summation of a constant is:

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

Step 3.2

Substitute the values into the formula.

$(- 7) (12)$

Step 3.3

Multiply $- 7$ by $12$ .

$- 84$

Step 4

Add the results of the summations.

$1300 - 84$

Step 5

Subtract $84$ from $1300$ .

$1216$