How Does the Sum of Alternating Series Lead to a Power Series for 2/(1-x^2)?

[kdinser]

Find a power series for the function centered at c and determine the interval of convergence.
c = 0

f(x)=\frac{2}{1-x^2}

After some partial fractions work and getting the partials in the form of

\frac{a}{1-r}

I have

\sum x^n + \sum(-x)^n

if I factor out the x^n's I get

\sum(1+(-1)^n)x^n

This is where I'm stuck, the solution manual shows it then going to

\sum2x^{2n}

I've been staring at this thing for 15 mins and can't see how it's possible. Could someone give me a little push in the right direction with this? Thanks.

 

[Dr Transport]

write out th efirst few terms and you will see the answer, n = 0, 1+1 = 2; n = 1, (1 + (-1))x = 0, etc...,

 

[marlon]

easy...

n has to be even so that the coefficient wouldn't be zero. Then you have 1+1 = 2 and the exponent is written as 2n to make sure it is even...



marlon

 

[kdinser]

Thanks guys, I wish the book or the manual would make it a little clearer when they do stuff like this. The way it's written seems to imply that there's some algebra going on to get to the final result.