Precalculus Examples

$\sum_{K = 1}^{7} - 3 {(2)}^{k - 1}$

Step 1

Factor $- 3$ out of the summation.

$(- 3) \sum_{K = 1}^{7} 2^{k - 1}$

Step 2

The formula for the summation of a constant is:

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

Step 3

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

$(- 3) ((2^{k - 1}) (7))$

Step 4

Simplify.

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Step 4.1

Move $7$ to the left of $2^{k - 1}$ .

$- 3 (7 \cdot 2^{k - 1})$

Step 4.2

Multiply $7$ by $- 3$ .

$- 21 \cdot 2^{k - 1}$