Calculus Examples

\sum_{i = 1}^{16} 5 i - 4

$\sum_{i = 1}^{16} 5 i - 4$

Step 1

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

\sum_{k = 1}^{16} 5 i - 4 = 5 \sum_{k = 1}^{16} i + \sum_{k = 1}^{16} - 4

$\sum_{k = 1}^{16} 5 i - 4 = 5 \sum_{k = 1}^{16} i + \sum_{k = 1}^{16} - 4$

Step 2

Evaluate

5 \sum_{k = 1}^{16} i

$5 \sum_{k = 1}^{16} i$ .

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Step 2.1

The formula for the summation of a polynomial with degree

1

$1$ is:

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

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

Step 2.2

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

(5) (\frac{16 (16 + 1)}{2})

$(5) (\frac{16 (16 + 1)}{2})$

Step 2.3

Simplify.

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Step 2.3.1

Add

16

$16$ and

1

$1$ .

5 \frac{16 \cdot 17}{2}

$5 \frac{16 \cdot 17}{2}$

Step 2.3.2

Multiply

16

$16$ by

17

$17$ .

5 (\frac{272}{2})

$5 (\frac{272}{2})$

Step 2.3.3

Divide

272

$272$ by

2

$2$ .

5 \cdot 136

$5 \cdot 136$

Step 2.3.4

Multiply

5

$5$ by

136

$136$ .

680

$680$

680

$680$

680

$680$

Step 3

Evaluate

\sum_{k = 1}^{16} - 4

$\sum_{k = 1}^{16} - 4$ .

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Step 3.1

The formula for the summation of a constant is:

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

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

Step 3.2

Substitute the values into the formula.

(- 4) (16)

$(- 4) (16)$

Step 3.3

Multiply

- 4

$- 4$ by

16

$16$ .

- 64

$- 64$

- 64

$- 64$

Step 4

Add the results of the summations.

680 - 64

$680 - 64$

Step 5

Subtract

64

$64$ from

680

$680$ .

616

$616$