Question

solve for x:

Answer 1

where the number

Answer 2

What's the question?

Answer 3

\[\sum_{n=1}^{nifty} (2x)^(11n) =44\]

Answer 4

11n is the power

Answer 5

nifty?

Answer 6

it looks like you try to write (2x)^(11n) ?

Answer 7

yeah, sorry. I'm not used to the formatting. It's a sum from n=1 to infinity of 2x raised to the power 11n equal to 44

Answer 8

\[\sum_{n=1}^{\infty} 2x^{11n}=44 \text{ or } \sum_{n=1}^{\infty} (2x)^{11n}=44\]
either way we will use geometric series thingy formula :p

Answer 9

Yeah, so that's what I was thinking. Divide the 2 over and you get x^(11n) = 22 and the sum of a geometric series = a1/(1-r) where r is the common ratio. I know that a geometric series only converges (and has a sum) if the common ratio r < 1.

Answer 10

\[\sum_{n=1}^{\infty} r^{k}=\frac{r}{1-r}\]

Answer 11

where r would be ?

Answer 12

x ... ?

Answer 13

That's the part where I was confused.

Answer 14

are there any restrictions on the value of x?

Answer 15

yeah, x < 1

Answer 16

if x= r

Answer 17

i mean in the question

Answer 18

oops sorry didnt read the question correctly

Answer 19

oh, no. Not that I'm aware of. I included all of the information about the question

Answer 20

almost
\[2 \sum_{n=1}^{\infty} x^{11n}=2 \sum_{n=1}^{\infty} (x^{11})^n\]

Answer 21

x^(11) should be r

Answer 22

oh okay.

Answer 23

also I messed up my formula earlier
I put n's and k's
:p
should be n's or k's not both
\[\sum_{n=1}^{\infty} r^{n}=\frac{r}{1-r} \\ \sum_{n=1}^{\infty} (x^{11})^n=\frac{?}{1-?}\]

Answer 24

Alright, I got x = (22/23)^(1/11) and it was correct. Thanks so much @freckles :)

Answer 25

np :)

Answer 26

Oh wow you are a smarty @freckles