Precalculus Examples

$\sum_{0}^{28} {(- 1)}^{n} \cdot (2 n)$

Step 1

Split the summation to make the starting value of $n$ equal to $1$ .

$\sum_{0}^{28} {(- 1)}^{n} \cdot (2 n) = \sum_{n = 1}^{28} {(- 1)}^{n} \cdot (2 n) + \sum_{n = 0}^{0} {(- 1)}^{n} \cdot (2 n)$

Step 2

Evaluate $\sum_{n = 1}^{28} {(- 1)}^{n} \cdot (2 n)$ .

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Step 2.1

Factor $2$ out of the summation.

$(2) \sum_{n = 1}^{28} {(- 1)}^{n} n$

Step 2.2

Expand the series for each value of $n$ .

${(- 1)}^{1} \cdot (2 \cdot 1) + {(- 1)}^{2} \cdot (2 \cdot 2) + {(- 1)}^{3} \cdot (2 \cdot 3) + \dots + {(- 1)}^{28} \cdot (2 \cdot 28)$

Step 2.3

Simplify the expanded form.

$28$

Step 3

Evaluate $\sum_{n = 0}^{0} {(- 1)}^{n} \cdot (2 n)$ .

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Step 3.1

Expand the series for each value of $n$ .

${(- 1)}^{0} \cdot (2 \cdot 0)$

Step 3.2

Simplify.

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Step 3.2.1

Raise $- 1$ to the power of $0$ .

$1 \cdot (2 \cdot 0)$

Step 3.2.2

Multiply $2$ by $0$ .

$1 \cdot 0$

Step 3.2.3

Multiply $1$ by $0$ .

$0$

Step 4

Replace the summations with the values found.

$28 + 0$

Step 5

Add $28$ and $0$ .

$28$