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

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The discussion focuses on simplifying the infinite series \(\sum^{\infty}_{n = 1} \frac{4^n}{3^{n-1}}\) into the form \(ar^{n-1}\). The transformation involves recognizing that \(4^n\) can be expressed as \(4^{1+(n-1)}\), which simplifies to \(4 \cdot 4^{n-1}\). This allows the series to be rewritten as \(\sum^{\infty}_{n = 1} 4\left(\frac{4}{3}\right)^{n-1}\). The confusion arises from understanding where the extra factor of 4 comes from and how the exponent manipulation works. Ultimately, the algebraic steps clarify the series' conversion into the desired format.
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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 :confused:
 
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shamieh said:
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 :confused:

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

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