How did they reduce this down to 1/n

from
[4^(n)/n] [(-1/2)^(2n)]

to
1/n

thanks!

[(4^n)/n] [(-1/2)^(2n)]

Let's do this step by step.
First, 4 is the same as 2^2.
[ ((2^2)^n)/n ] [(-1/2)^(2n)]

Next, whenever we have a power to a power, we can multiply the powers, i.e. (a^b)^c = a^(bc).
Multiplying out the 2 and the n, we get

[(2^(2n))/n] [ (-1/2)^(2n) ]

Next, we are going to use the property that (a/b)^m = (a^m)/(b^m).

[(2^(2n))/n] [ ( (-1)^(2n) )/2^(2n) ]

We're going to do with the (-1) that we did with the 4, except backwards. We're going to express (-1)^(2n) as [(-1)^2]^n.

[(2^(2n))/n] [ ((-1)^2)^n / 2^(2n) ]

And (-1) squared is equal to 1.

[(2^(2n))/n] [ (1^n / 2^(2n) ]

1 to the power of anything is 1.

[(2^(2n))/n] [ (1 / 2^(2n) ]

We have a cross-cancellation, as we are multiplying two fractions. We are left, after the cancellation, with

[1/n] [1/1]

Which is just

1/n

Is n restricted to integer values? If so, consider the following......

4^n=(2*2)^2=(2^2)^n=2^(2n)

And (-1/2)^(2n)=(-1)^(2n)/(2^(2n)) If n is restricted to integer values then (-1)^(2n) always equals 1, because you are raising negative one to an even power.

So you have....

(2^(2n))/(n(2^(2n))=1/n

If n can be any real number, then it becomes (-1)^(2n)/n

We have to assume n is an integer, otherwise (-1/2)^(2n) may not be defined.

(-1/2)^(2n) = ((-1/2)^2)^n = (1/4)^n.

Thus, we have (4^n / n) * (1/4)^n = (4 * 1/4)^n / n = 1^n / n = 1/n.