# How do you write the rule for the nth term given #2,-4,8,-10,14,...?

Nov 5, 2016

${a}_{n} = 2 + {\left(- 1\right)}^{n} 6 \left(1 - n\right)$

#### Explanation:

Given:

$2 , - 4 , 8 , - 10 , 14$

Look at the sequence of differences between successive terms:

$- 6 , 12 , - 18 , 24$

The absolute values of these are in arithmetic progression with common difference $6$, but with alternating signs.

Hence we can express our original sequence by a recursive rule:

$\left\{\begin{matrix}{a}_{1} = 2 \\ {a}_{n + 1} = {a}_{n} + {\left(- 1\right)}^{n} 6 n\end{matrix}\right.$

How about a general formula for ${a}_{n}$ in terms of $n$?

Looking at the alternate terms we can observe that we have two interwoven arithmetic sequences - one ($2 , 8 , 14 , \ldots$) with a common difference of $6$ and the other ($- 4 , - 10 , - 16 , \ldots$) with a common difference of $- 6$.

Hence we can write:

${a}_{n} = \left\{\begin{matrix}6 n - 4 \text{ if " n " is odd" \\ 8-6n " if " n " is even}\end{matrix}\right.$

Note that if we subtract $2$ from each of the expressions for odd and even $n$ then we can make the expressions look a little more similar:

${a}_{n} = \left\{\begin{matrix}2 - \left(6 - 6 n\right) \text{ if " n " is odd" \\ 2+(6-6n) " if " n " is even}\end{matrix}\right.$

and hence:

${a}_{n} = 2 + {\left(- 1\right)}^{n} 6 \left(1 - n\right)$