Q.Find the coefficient of $x^n$ in the expansion of $\dfrac1{(1-x)(1-2x)}$ as a series of ascending powers of $x$, stating the range of validity.

Step 1. Resolve $\dfrac1{(1-x)(1-2x)}=\dfrac A{1-x}+\dfrac B{1-2x}$: clearing denominators, $1=A(1-2x)+B(1-x)$.

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

Step 3. Substitute $x=\dfrac12$: $1=B\big(1-\dfrac12\big)=\dfrac B2\ \Rightarrow\ B=2$.

Step 4. So $\dfrac1{(1-x)(1-2x)}=\dfrac{-1}{1-x}+\dfrac2{1-2x}$.

Step 5. Expand each term as a geometric series: $\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$ (valid $|x|<1$), and $\dfrac1{1-2x}=\displaystyle\sum_{n=0}^\infty (2x)^n=\sum_{n=0}^\infty 2^n x^n$ (valid $|x|<\dfrac12$).

Step 6. So the coefficient of $x^n$ in the sum is $-1+2\cdot2^n=2^{n+1}-1$.

Step 7 (validity). The series for $\dfrac1{1-x}$ needs $|x|<1$ and the series for $\dfrac1{1-2x}$ needs $|x|<\dfrac12$; the whole expansion is valid only where both hold, i.e. $|x|<\dfrac12$.

Step 8 (check, $n=0$). Coefficient should be $f(0)=1$; formula gives $2^1-1=1$. Matches.

Partial fractions into two non-repeated-linear-factor terms, each expanded as a geometric series; validity is the intersection of the two series' radii of convergence.

• Stating the validity as $|x|<1$ (the radius of the easier term) instead of the more restrictive $|x|<\dfrac12$.
• Sign slip in $A(1-2x)+B(1-x)=1$ when substituting $x=1$.