Q.Find the coefficient of $x^3$ in the expansion of $\dfrac{2x+3}{(1-x)(2+x)}$ as a series of ascending powers of $x$, stating the range of validity.

Step 1. Resolve $\dfrac{2x+3}{(1-x)(2+x)}=\dfrac A{1-x}+\dfrac B{2+x}$: clearing denominators,

Step 2. Substitute $x=1$: $5=3A\ \Rightarrow\ A=\dfrac53$.

Step 3. Substitute $x=-2$: $-1=3B\ \Rightarrow\ B=-\dfrac13$.

Step 4. So $\dfrac{2x+3}{(1-x)(2+x)}=\dfrac{5/3}{1-x}-\dfrac{1/3}{2+x}$.

Step 5. Rewrite the second term with $2+x=2\big(1+\dfrac x2\big)$: $-\dfrac{1/3}{2(1+x/2)}=-\dfrac16\cdot\dfrac1{1+x/2}=-\dfrac16\sum_{n=0}^\infty\Big(\!-\dfrac x2\Big)^n=-\dfrac16\sum_{n=0}^\infty(-1)^n\Big(\dfrac x2\Big)^n$.

Step 6. The first term expands as $\dfrac53\displaystyle\sum_{n=0}^\infty x^n$.

Step 7. Coefficient of $x^n$: $\dfrac53-\dfrac16\cdot\dfrac{(-1)^n}{2^n}=\dfrac53-\dfrac{(-1)^n}{6\cdot2^n}$.

Step 8. Substitute $n=3$: $\dfrac53-\dfrac{(-1)^3}{6\cdot8}=\dfrac53+\dfrac1{48}=\dfrac{80}{48}+\dfrac1{48}=\dfrac{81}{48}=\dfrac{27}{16}$. …

Partial fractions with one factor rewritten in the form $1+kx$ before expanding as a geometric series, then eval …

• Forgetting to rewrite $2+x$ as $2(1+x/2)$ before expanding, which is needed to get the series in the standard $\sum(\pm x/2)^n$ form. …