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

Aug 2, 2016

${a}_{6} = 4096$ or ${a}_{6} = - 1024$ depending on convention.

#### Explanation:

The series here looks to be:
${\sum}_{n = 0} {\left(- 4\right)}^{n}$

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:
${a}_{n} = {\left(- 4\right)}^{n}$

However if we want to start counting from $n = 1$ then we can write the formula as:
${a}_{n} = {\left(- 4\right)}^{n - 1}$

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.