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

1 Answer
Nov 5, 2016

Answer:

#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)#