Series Simplification: Simplifying \sum^{\infty}_{n = 1} 4^n/3^{n-1}

[shamieh]

They are re writing it to be in the form $$ar^{n-1}$$ But I can't figure out how

$$\sum^{\infty}_{n = 1} \frac{4^n}{3^{n- 1}}$$ turns into this: $$\sum^{\infty}_{n = 1} 4(\frac{4}{3})^{n-1}$$ ??

I'm confused on the Algebra they are implementing to do this..Can anyone elaborate? For example, where is the top n, where is the extra 4 coming from.. I know this is simple algebra but It's flying right over me

 

[SuperSonic4]

They are re writing it to be in the form $$ar^{n-1}$$ But I can't figure out how

$$\sum^{\infty}_{n = 1} \frac{4^n}{3^{n- 1}}$$ turns into this: $$\sum^{\infty}_{n = 1} 4(\frac{4}{3})^{n-1}$$ ??

I'm confused on the Algebra they are implementing to do this..Can anyone elaborate? For example, where is the top n, where is the extra 4 coming from.. I know this is simple algebra but It's flying right over me

From the laws of exponents you can say $$a^{b+c} = a^b\, a^c$$

In this case you can write $$4^n = 4^{1+(n-1)} = 4^1 \cdot 4^{n-1}$$