Here's the question you clicked on:
wcaprar
What is the sum of the infinite series?
\[\sum_{n=2}^{\infty} 2^n/3*5^{n+2}\]
I see that you can take out the 5^2 and multiply that with three then simplify to get \[(1/75)*(2/5)^n\] which is a geometric series. But when I solve it out like you would a regular geometric series I get 1/45. But wolframalpha says 20
your idea is correct. lets make sure the arithmetic is correct
\[\frac{1}{75}\sum_{n=2}^{\infty}\left(\frac{2}{5}\right)^n=\frac{1}{75}\left(\frac{\frac{4}{625}}{1-\frac{2}{5}}\right)\]
How do you get 4/625 on top?
we are starting at n=2, not n=1. The first term of the sequence is 4/625.