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

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

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

Step 3. Substitute $x=\dfrac12$: $\dfrac12=C\big(1-\dfrac12\big)^2=\dfrac C4\ \Rightarrow\ C=2$.

Step 4. Compare the coefficient of $x^2$. LHS: $0$. RHS: $A(1-x)(1-2x)=A(1-3x+2x^2)$ contributes $2A$; $C(1-x)^2=C(1-2x+x^2)$ contributes $C$; the $B$ term contributes none. So $0=2A+C=2A+2\ \Rightarrow\ A=-1$.

Step 5 (check, constant term). LHS: $0$. RHS: $A+B+C=-1-1+2=0$. Matches.

Step 6. So $\dfrac{x}{(1-x)^2(1-2x)}=\dfrac{-1}{1-x}+\dfrac{-1}{(1-x)^2}+\dfrac2{1-2x}$.

Step 7. Expand each term: $\dfrac1{1-x}=\displaystyle\sum x^n$; $\dfrac1{(1-x)^2}=\displaystyle\sum(n+1)x^n$ (the derivative form of the geometric series); $\dfrac1{1-2x}=\displaystyle\sum2^nx^n$.

Step 8. Coefficient of $x^n$ is $-1-(n+1)+2\cdot2^n=2^{n+1}-n-2$.

Step 9 (validity). The three series require $|x|<1$, $|x|<1$, $|x|<\dfrac12$ respectively; the tightest is $|x|<\dfrac12$.

Step 10 (check, $n=0,1$). $n=0$: $2^1-0-2=0$ (correct, since $f(x)=x\cdot(\cdots)$ starts at $x^1$). $n=1$: $2^2-1-2=1$, matching $f(x)\approx x$ for small $x$ (since the bracketed factor $\to1$ as $x\to0$).

Partial fractions with a repeated linear factor, expanded term-by-term using the geometric series and its derivative form $1/(1-x)^2=\sum(n+1)x^n$; validity is the intersection of the radii of convergence.

• Using $\sum x^n$ for the repeated factor $1/(1-x)^2$ instead of the correct $\sum(n+1)x^n$.
• Missing the sanity check that the coefficient of $x^0$ must be $0$, since the original function is $x$ times a power series.