Question

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

Answer 1

that was a good effort :)

Answer 2

gonna have to prolly rewrite it tho

Answer 3

is this \[\sum3^{\frac{2}{3}}(n+1)\]?

Answer 4

no probably not because that would be infinite

Answer 5

http://gracecatholicschool.org/AngelThemes/MML/2010/10/19/151250_726268.gif

Answer 6

ahhhhk a geometric series! ignore the first 3 or rather pull it out front of the summation

Answer 7

6

Answer 8

just 6?

Answer 9

yeah

Answer 10

actually i think it is 4

Answer 11

soooo 4?

Answer 12

sorry 4...lol

Answer 13

0 or 4?

Answer 14

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

Answer 15

4

Answer 16

change n + 1 to n and start at n = 0

Answer 17

.... i change my vote to 'im an idiot' :)

Answer 18

ok so 4 it is

Answer 19

it is 4 if you just want the answer

Answer 20

yesssss

Answer 21

thnx

Answer 22

NO THANKK YOUUUU :)

Answer 23

actually satellite73, if you change n+1 to n you would have to start from n=2