# Find an equation for the nth term of the sequence. {eq}-2, -8, -32, -128, ... {/eq}

## Question:

Find an equation for the nth term of the sequence.

{eq}-2, -8, -32, -128, ... {/eq}

## General Term of a Sequence:

A sequence is a function {eq}f:\mathbb{N} \to \mathbb{R} {/eq} i.e., its domain is the set of natural number.

Suppose {eq}\{a_n\} {/eq} is a given sequence, then {eq}a_n {/eq} is known as its general term.

From general term we can find any term of the sequence.

Explanation:

Given that, first four terms of the sequence {eq}-2, -8, -32, -128, ... {/eq}

Since, all term are negative, so {eq}a_n {/eq} is also negative.

Also, these all terms are the odd powers of -2.

i.e., {eq}(-2)^1, (-2 )^{3}, (-2)^5, (-2)^7...... {/eq}

So, the general term is {eq}a_n = (-2)^{2n-1}. {/eq}

Conclusion:

The sequence is {eq}\{a_n\} = \{(-2)^{2n-1} \}. {/eq}