What is the reciprocal of $(-1/2)^k$?

Question

What is the reciprocal of $(-1/2)^k$?

The answer is meant to be $2^{-k}$ as if you flip something upside down the power becomes negative. However, I am not sure what happens to the negative in front of the fraction.

Answer 1

The answer would look like $(-1)^k.2^{-k}$. The negative would depend if k is even or odd.

Answer 2

A series of simple rewrites using very basic facts aout exponentiation gives: $$ \left(-\frac12\right)^k =\left(\frac{-1}{2}\right)^k =\frac{(-1)^k}{2^k} =(-1)^k2^{-k} $$

If $k$ is even $(-1)^k$ is $1$, giving the result you say it is meant to give.

Answer 3

The reciprocal of $x$ is simply $\frac1x$. So the reciprocal of $(-\tfrac12)^k$ is $\frac1{(-\tfrac12)^k}$.

You may also recall from the exponentiation laws that $(\frac1x)^y=x^{-y}$, and that the repciprocal of $x^y$ is $\frac1{x^y}=x^{-y}$. This allows us to rewrite $(-\tfrac 12)^k$ as $(-2)^{-k}$, and then the reciprocal of this is $(-2)^k$. If $k$ is even, this simplifies to $2^k$; if $k$ is odd, it becomes $-2^k$.

Answer 4

Condensed form of Hagen's answer: $$ \frac{1}{(-\frac{1}{2})^k}=\frac{1}{\frac{(-1)^k}{2^k}}=\frac{2^k}{(-1)^k}=\begin{cases}2^k &\text{if $k$ is even,}\\ -2^k &\text{if $k$ is odd.}\end{cases} $$