Simple Algebra Problem

[ku]

How does [math]\frac{4^n}{3^{n-1}}=4\left(\frac{4}{3}\right)^{n-1}[/math]?

[Dave]

It's fairly simple;

 

4ⁿ/3^n-1 = 4*4^-1 * 4ⁿ/3^n-1

 

So we have: 4*4^n-1/3^n-1 = 4(4/3)^n-1.