Question

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

Answer 1

`a_n = 2+(-1)^n6(1-n)`

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:

`{ (a_1 = 2), (a_(n+1) = a_n +(-1)^n 6n) :}`

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,...`) with a common difference of `6` and the other (`-4, -10, -16,...`) with a common difference of `-6`.

Hence we can write:

`a_n = { (6n-4 " if " n " is odd"), (8-6n " if " n " is even") :}`

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 = { (2-(6-6n) " if " n " is odd"), (2+(6-6n) " if " n " is even") :}`

and hence:

`a_n = 2+(-1)^n6(1-n)`