# Another infinite sequence

• March 15th 2009, 10:54 PM
mollymcf2009
Another infinite sequence
Last one I'm stuck on:

$a_n$ = ?

{ $1, -\frac{2}{3}, .4444444444444444, - .296296296296296296,.....$}
• March 15th 2009, 10:57 PM
matheagle
Quote:

Originally Posted by mollymcf2009
Last one I'm stuck on:

$a_n$ = ?

{ $1, -\frac{2}{3}, .4444444444444444, - .296296296296296296,.....$}

$(-2/3)^n$ where n=0,1,2,...

or

$(-2/3)^{n-1}$ where n=1,2,3...
• March 15th 2009, 11:07 PM
Quote:

Originally Posted by matheagle
$(-2/3)^n$ where n=0,1,2,...

or

$(-2/3)^{n-1}$ where n=1,2,3...

Can you explain this for the fourth term its not coming correctly (Thinking)

EDIT: It came thanks (Yes)
• March 15th 2009, 11:13 PM
Quote:

Originally Posted by mollymcf2009
Last one I'm stuck on:

$a_n$ = ?

{ $1, -\frac{2}{3}, .4444444444444444, - .296296296296296296,.....$}

Here is the explanation about third and 4th term

$x = .4444444444444444.. ---> (1)$

$10x = 4.44444...----->(2)$

(2) - (1)

$9x= 4$

$
x = \frac{4}{9} = \frac{ 2\times 2}{ 3 \times 3}$

$x = -0.296296296296296296 ..---> (1)$

$1000x =- 296.296296......----> (2)$

(2)-(1)

$999x = -296$

$x = \frac{-296}{999} = \frac{4 \times \not 37 \times -2}{9 \times \not 37 \times 3 }$
• March 15th 2009, 11:17 PM
matheagle
That's way too much work 4 me.
I just used my calculator to check (2/3) to the third power.
The first three are obvious.