# How do you Find the n-th term of the infinite sequence 1,-2/3,4/9,-8/27,…?

Jul 31, 2014

When looking at sequences, your mind should always be in a pattern-recognizing mode. Sometimes sequences seem daunting, but when you look at them in pieces, they're usually easier to tackle.

With this sequence, we can see that for each term, the numerator doubles, and the denominator triples. Also, it's important to note that each successive term flips sign. (positive to negative to positive)

To tackle the sign flips, we know we'll have to put in our formula a term raising $- 1$ to a power involving $n$. Since the sign flips happen every odd term in the sequence, the term in our formula will need to look something like this:

${\left(- 1\right)}^{n - 1}$

When $n$ is 1, this thing will equate out to ${\left(- 1\right)}^{0}$, which is just $1$. When $n$ is 2, it'll be $- 1$, and so on. In our general formula, we'll multiply this bit against everything else, to cause the sign flips.

Since each successive numerator is double the previous one, we can use ${2}^{n - 1}$ in the formula's numerator. At $n = 1$, this will be $1$, at $n = 2$, it'll be $2$, at $n = 3$, it'll be $4$, and so on.

For the denominator, we'll use ${3}^{n - 1}$ This results in $1 , 3 , 9 ,$ etc..

Now, we simply piece them together:

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