the limiting sum of 3 + x + x^2 + ..... is 18. If |x|<1, find x

how do you do this??

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- Dec 7th 2009, 08:03 PMdeej813limiting sum of a series??
the limiting sum of 3 + x + x^2 + ..... is 18. If |x|<1, find x

how do you do this?? - Dec 7th 2009, 08:20 PMalexmahone
- Dec 7th 2009, 08:27 PMdeej813
how does that work??

where does the x/1-x come from?? - Dec 7th 2009, 09:35 PMProve It
Everything after the 3 is an infinite geometric series.

For any infinite geometric series of the form

$\displaystyle S_n = a + ar + ar^2 + ar^3 + \dots$

converges to $\displaystyle \frac{a}{1 - r}$, provided $\displaystyle |r| < 1$.

In your case, the infinite geometric series is

$\displaystyle S_n = x + x^2 + x^3 + x^4 + \dots$.

Can you see what $\displaystyle a$ and $\displaystyle r$ have to be?

Can you see what it converges to if $\displaystyle |x| < 1$? - Dec 7th 2009, 09:46 PMdeej813
wouldn't it be 3/1-x cos isn't a=3??

- Dec 7th 2009, 09:56 PMProve It
No.

I said the terms AFTER the 3.

In other words, your equation is

$\displaystyle 3 + \left(x + x^2 + x^3 + \dots\right) = 18$.

We are only dealing with the stuff inside the brackets.

Edit: It might help if you subtract the 3 to the other side.

So the equation becomes

$\displaystyle x + x^2 + x^3 + \dots = 15$. - Dec 7th 2009, 10:01 PMdeej813
oh ok i get it

thanks :)