# What would the formula for the nth term be given -2, 6, -18, 54...?

Feb 16, 2016

$f \left(n\right) = - 2 \cdot {\left(- 3\right)}^{n - 1}$

#### Explanation:

The easiest way to explain this is to say: The first term is $- 2$ and each number in the sequence after that is the previous one times $- 3$. That is called a recursive definition and would look like this:
$f \left(1\right) = - 2$
$f \left(n\right) = F \left(n - 1\right) \cdot - 3$

Once we start with $f \left(1\right)$ we can just plug it into the second line to get $f \left(2\right) = 6$, $f \left(3\right) = - 18$, and so on. But if we want a single equation that will let us know $f \left(100\right)$ without calculating all 99 previous numbers in the sequence, that formula is pretty much useless.

$f \left(2\right) = - 2 \cdot - 3 = - 2 \cdot {\left(- 3\right)}^{1}$
$f \left(3\right) = - 2 \cdot - 3 \cdot - 3 = - 2 \cdot {\left(- 3\right)}^{2}$
$f \left(4\right) = - 2 \cdot - 3 \cdot - 3 \cdot - 3 = - 2 \cdot {\left(- 3\right)}^{3}$
Notice that the exponent that $- 3$ is getting raised to is 1 less than the sequence number we want. It gets raised to 1 when $n$ is 2, and so on. That also works in our favor as ${x}^{0} = 1$ so that $f \left(1\right) = - 2 \cdot {\left(- 3\right)}^{0}$.
$f \left(n\right) = - 2 \cdot {\left(- 3\right)}^{n - 1}$