Find coefficient on power series

[jjangub]

(Moderator's note: thread moved from "General Math")

Hi.
I am confused with this question. I tried two different ways to solve it, but I got different answers for each way. The question is
"Determine the coefficient of x^100 in the pwer series form of (1+x+x^2)/((1-x^3)^2)"
First, I tried this way,
since we know 1/(1-x^3)=(1+x^3+x^6+x^9+...), so the form becomes
(1+x+x^2)*(1+x^3+x^6+x^9+...)^2
let a0,a1 be the power of x from each series,then a0+a1=100.
from the first part we know that a0=0 or 1 or 2.
When a0=0, a1^2=a1+a1=100
when a0=1, a1^2=a1+a1=99
when a0=2, a1^2=a1+a1=98
and we know that a1 is multiple of 3, let a1=3n, then 6n=100.
Since we do not have n value, the coefficient of x^100 is 0.
Second I tried this way,
since we know 1+x+x^2=(1-x^3)/(1-x), if we apply this to our form, (1-x^3) from top and bottom is gone, therefore we have (1/(1-x))*(1/(1-x^3)).
since we know, 1/(1-x)=1+x+x^2+x^3+...
and 1/(1-x^3)=1+x^3+x^6+x^9+...
therefore we now have (1+x+x^2+x^3+...)*(1+x^3+x^6+x^9+...)
let a0 and a1 be the power of x from each series, then a0+a1=100.
Since a1 is the multiple of 3, it can be written as 3a1, then a0+3a1=100.
We have (a0,a1) as (1,33),(10,30),(19,27),(28,24)...(100,0) there are 12 sets
and each has coefficient of 1,
therefore the coefficient is 12.
I got two different answers. I don't know what is the problem...
Please tell me whether I am right or wrong...

 

[mathman]

Both calculations are incorrect!
For the first approach when a0=1, the following a1 pairs are allowed (0,99), (3,96),...(96,3), (99,0). 34 pairs.
For the second set, you need the following (a0,a1) pairs (1,99), (4,96),...(97,3), (100,0). 34 pairs.