shamieh
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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
$$\sum^{\infty}_{n = 1} \frac{4^n}{3^{n- 1}}$$ 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
