Question

One sec.

Answer 1

PONYYYYYYYYYYYYYYYYY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

\[\sum_{n=1}^{\infty}(3/8)(3/4)^{n-1}\]

Answer 3

first off pull the \(\frac{3}{8}\) right out front

Answer 4

\[\frac{3}{8}\sum_{n=1}^{\infty}\left(\frac{3}{4}\right)^{n-1}\]

Answer 5

okay. what does that do?

Answer 6

makes it 20% easier.

Answer 7

it allows you to use the formula for summing a geometric series
\[\sum_{k=1}^{\infty}r^{k-1}=\frac{1}{1-r}\]

Answer 8

Alright. I wish i could give all of you a medal, this has made my day.

Answer 9

in your case \(r=\frac{3}{4}\)

Answer 10

with some practice you can do it in your head
one minus three fourths is one fourth
the reciprocal of one fourth is 4

at the end multiply \(4\times \frac{3}{8}\) because of that pesky \(\frac{3}{8}\) out front

Answer 11

Ohh mulitply by 3/8 too. thanks :)

Answer 12

yeah that one is still out front, you can't forget about it

Answer 13

Cool. Thanks!

Answer 14

yw