this step in the proof is confusing me:
$$\sum_1^\infty {\frac{4^{n}}{3^{n-1}}}\qquad \longrightarrow \qquad\sum_1^\infty 4\left(\frac43\right)^{n-1}$$
please explain how/why this happened?
cheers,
gregg
this step in the proof is confusing me:
$$\sum_1^\infty {\frac{4^{n}}{3^{n-1}}}\qquad \longrightarrow \qquad\sum_1^\infty 4\left(\frac43\right)^{n-1}$$
please explain how/why this happened?
cheers,
gregg
The $n^\text{th}$ term was rewritten by pulling out a factor of $4$ from the numerator. Maybe seeing a couple of extra steps will help:
$$\frac{4^n}{3^{n-1}}=\frac{4\cdot4^{n-1}}{3^{n-1}}=4\frac{4^{n-1}}{3^{n-1}}=4\left(\frac{4}{3}\right)^{n-1}$$