Calculus Examples

$\sum_{j = 1}^{15} {(j - 1)}^{3}$

Step 1

Simplify the summation.

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Step 1.1

Use the Binomial Theorem.

$j^{3} + 3 j^{2} \cdot - 1 + 3 j {(- 1)}^{2} + {(- 1)}^{3}$

Step 1.2

Simplify each term.

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Step 1.2.1

Multiply $- 1$ by $3$ .

$j^{3} - 3 j^{2} + 3 j {(- 1)}^{2} + {(- 1)}^{3}$

Step 1.2.2

Raise $- 1$ to the power of $2$ .

$j^{3} - 3 j^{2} + 3 j \cdot 1 + {(- 1)}^{3}$

Step 1.2.3

Multiply $3$ by $1$ .

$j^{3} - 3 j^{2} + 3 j + {(- 1)}^{3}$

Step 1.2.4

Raise $- 1$ to the power of $3$ .

$j^{3} - 3 j^{2} + 3 j - 1$

Step 1.3

Rewrite the summation.

$\sum_{j = 1}^{15} j^{3} - 3 j^{2} + 3 j - 1$

Step 2

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

$\sum_{j = 1}^{15} j^{3} - 3 j^{2} + 3 j - 1 = \sum_{j = 1}^{15} j^{3} - 3 \sum_{j = 1}^{15} j^{2} + 3 \sum_{j = 1}^{15} j + \sum_{j = 1}^{15} - 1$

Step 3

Evaluate $\sum_{j = 1}^{15} j^{3}$ .

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Step 3.1

The formula for the summation of a polynomial with degree $3$ is:

$\sum_{k = 1}^{n} k^{3} = \frac{n^{2} {(n + 1)}^{2}}{4}$

Step 3.2

Substitute the values into the formula.

$\frac{15^{2} {(15 + 1)}^{2}}{4}$

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 $15$ and $1$ .

$\frac{15^{2} \cdot 16^{2}}{4}$

Step 3.3.1.2

Raise $15$ to the power of $2$ .

$\frac{225 \cdot 16^{2}}{4}$

Step 3.3.1.3

Raise $16$ to the power of $2$ .

$\frac{225 \cdot 256}{4}$

Step 3.3.2

Simplify the expression.

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Step 3.3.2.1

Multiply $225$ by $256$ .

$\frac{57600}{4}$

Step 3.3.2.2

Divide $57600$ by $4$ .

$14400$

Step 4

Evaluate $3 \sum_{j = 1}^{15} j^{2}$ .

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Step 4.1

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 4.2

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

$(- 3) (\frac{15 (15 + 1) (2 \cdot 15 + 1)}{6})$

Step 4.3

Simplify.

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Step 4.3.1

Simplify the numerator.

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Step 4.3.1.1

Add $15$ and $1$ .

$- 3 \frac{15 \cdot 16 (2 \cdot 15 + 1)}{6}$

Step 4.3.1.2

Combine exponents.

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Step 4.3.1.2.1

Multiply $15$ by $16$ .

$- 3 \frac{240 (2 \cdot 15 + 1)}{6}$

Step 4.3.1.2.2

Multiply $2$ by $15$ .

$- 3 \frac{240 (30 + 1)}{6}$

Step 4.3.1.3

Add $30$ and $1$ .

$- 3 \frac{240 \cdot 31}{6}$

Step 4.3.2

Reduce the expression by cancelling the common factors.

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Step 4.3.2.1

Multiply $240$ by $31$ .

$- 3 (\frac{7440}{6})$

Step 4.3.2.2

Cancel the common factor of $3$ .

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Step 4.3.2.2.1

Factor $3$ out of $- 3$ .

$3 (- 1) \frac{7440}{6}$

Step 4.3.2.2.2

Factor $3$ out of $6$ .

$3 \cdot - 1 \frac{7440}{3 \cdot 2}$

Step 4.3.2.2.3

Cancel the common factor.

$3 \cdot - 1 \frac{7440}{3 \cdot 2}$

Step 4.3.2.2.4

Rewrite the expression.

$- 1 (\frac{7440}{2})$

Step 4.3.2.3

Simplify the expression.

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Step 4.3.2.3.1

Divide $7440$ by $2$ .

$- 1 \cdot 3720$

Step 4.3.2.3.2

Multiply $- 1$ by $3720$ .

$- 3720$

Step 5

Evaluate $3 \sum_{j = 1}^{15} j$ .

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Step 5.1

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

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

Step 5.2

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

$(3) (\frac{15 (15 + 1)}{2})$

Step 5.3

Simplify.

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Step 5.3.1

Add $15$ and $1$ .

$3 \frac{15 \cdot 16}{2}$

Step 5.3.2

Multiply $15$ by $16$ .

$3 (\frac{240}{2})$

Step 5.3.3

Divide $240$ by $2$ .

$3 \cdot 120$

Step 5.3.4

Multiply $3$ by $120$ .

$360$

Step 6

Evaluate $\sum_{j = 1}^{15} - 1$ .

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Step 6.1

The formula for the summation of a constant is:

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

Step 6.2

Substitute the values into the formula.

$(- 1) (15)$

Step 6.3

Multiply $- 1$ by $15$ .

$- 15$

Step 7

Add the results of the summations.

$14400 - 3720 + 360 - 15$

Step 8

Simplify.

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Step 8.1

Subtract $3720$ from $14400$ .

$10680 + 360 - 15$

Step 8.2

Add $10680$ and $360$ .

$11040 - 15$

Step 8.3

Subtract $15$ from $11040$ .

$11025$