Question

4+2{sum from 1 to infinity}4/(3^n)

equals

4+2 * (4/3)/(1-(1/3))

What work was done between the first and second step?? D:

Answer 1

\[4+2\sum_{1}^{\infty}4/3^n = 4+2(4/3)/(1-(1/3))\]

Answer 2

is this
\[4+2\sum_{n=1}^{\infty}(\frac{4}{3})^n\] or
\[4+2\sum_{k=1}^{\infty}\frac{4}{3^n}\]

Answer 3

Oops, sorry - still getting the hang of using the equation editor.

It's the second one . . .

Answer 4

ok then the 4 can come right out front and you have
\[4+8\sum_{n=1}^{\infty}\frac{1}{3^n}\]

Answer 5

ignoring the 4 and the 8 you have
\[\sum_{n=1}^{\infty}\frac{1}{3^n}\] and the formula for that is
\[\frac{\frac{1}{3}}{1-\frac{1}{3}}\] is that part clear or no?

Answer 6

^ Why is that?

Answer 7

hmmm it take a bit of explaining.
do you know how to sum a geometric series?

Answer 8

in general
\[a+ar+ar^2+ar^3+...=\sum_{k=1}^{\infty}ar^k =\frac{a}{1-r}\] if
\[0<r<1\]

Answer 9

Oh - never mind. I understand that part now :D

Answer 10

typo above, i should have written
\[\sum_{k=0}^{\infty} ar^k=\frac{a}{1-r}\]

Answer 11

Thank you! I'm sorry I was a little slow. :)