Q.Find the coefficient of $x^4$ in the expansion of $\dfrac{x^2}{(1-x)(1-x^2)}$ as a series of ascending powers of $x$, stating the range of validity.

Step 1. Factor the denominator: $(1-x)(1-x^2)=(1-x)(1-x)(1+x)=(1-x)^2(1+x)$. So the function is $f(x)=\dfrac{x^2}{(1-x)^2(1+x)}$.

Step 2. Resolve $\dfrac{x^2}{(1-x)^2(1+x)}=\dfrac A{1-x}+\dfrac B{(1-x)^2}+\dfrac C{1+x}$: clearing denominators,

Step 3. Substitute $x=1$: $1=B(2)\ \Rightarrow\ B=\dfrac12$.

Step 4. Substitute $x=-1$: $1=C(2)^2=4C\ \Rightarrow\ C=\dfrac14$.

Step 5. Compare the coefficient of $x^2$. LHS: $1$. RHS: $A(1-x^2)$ contributes $-A$; $C(1-x)^2=C(1-2x+x^2)$ contributes $C$; the $B$ term contributes none. So $1=-A+C=-A+\dfrac14\ \Rightarrow\ A=-\dfrac34$.

Step 6 (check, constant term). LHS: $0$. RHS: $A+B+C=-\dfrac34+\dfrac12+\dfrac14=0$. Matches.

Step 7. So $f(x)=\dfrac{-3/4}{1-x}+\dfrac{1/2}{(1-x)^2}+\dfrac{1/4}{1+x}$.

Step 8. Expand each term: $\dfrac1{1-x}=\sum x^n$; $\dfrac1{(1-x)^2}=\sum(n+1)x^n$; $\dfrac1{1+x}=\sum(-1)^nx^n$.

Step 9. Coefficient of $x^n$: $-\dfrac34+\dfrac12(n+1)+\dfrac14(-1)^n$. …

Factoring a hidden repeated linear factor before resolving into partial fractions, then evaluating …

• Not noticing that $1-x^2=(1-x)(1+x)$ makes $(1-x)$ a repeated factor of the full denominator, and hence assigning only a single term $A/(1-x)$ instead of the required $A/(1-x)+B/(1-x)^2$. …