How do you write an nth term rule for #1,-4,16,-64# and find #a_6#?

1 Answer
Aug 2, 2016

Answer:

#a_6 = 4096# or #a_6 = -1024# depending on convention.

Explanation:

The series here looks to be:
#sum_(n=0) (-4)^n#

This does present a slight ambiguity in the way the question was asked. That is, if we assume that we start counting from #n=0# then:
#a_n = (-4)^n#

However if we want to start counting from #n=1# then we can write the formula as:
#a_n = (-4)^(n-1)#

The later formula is advantageous in that the #n#th term actually corresponds to that #n#. That is, in the first equation, the third term is at #a_2# as opposed to #a_3#. In the second equation, the #n#th term is denoted simply by #a_n#.

Thus, it is not completely clear what the correct answer would be for #a_6#.

It may be:
#a_6 = 4096#
if we started counting from #n=0#

Or
#a_6 = -1024#
if we started counting from #n = 1#

We'll note that the first answer is actually the 6th term in the series, while the second is actually the 5th. If this problem were presented on homework I would most likely ask for clarification from the instructor.