Question

Expand $\frac{7+x}{(1+x)(1+x^2)}$ in ascending powers of $x$ and find the general term.

Answer 1

Assume $\frac{7+x}{(1+x)(1+x^2)}=\frac{A}{1+x}+\frac{Bx+C}{1+x^2}$
$\therefore$ $7+x=A(1+x^2)+(Bx+C)(1+x)$
Let $1+x=0$, then $A=3$;
Equating the absolute terms, $7=A+C$, whence $C=4$
Equating the coefficients of $x^2$, $0=A+B$, whence $B=-3$
$\therefore \frac{7+x}{(1+x)(1+x^2)}=\frac{3}{1+x}+\frac{4-3x}{1+x^2}$
$={3(1+x)}^{-1}+(4-3x){(1+x^2)}^{-1}$
$=3\{1-x+x^2-......+{(-1)}^p{x}^p+....\}+(4-3x)\{1-x^2+x^4-......+{(-1)}^p{x}^{2p}+....\}$

To find the coefficient of $x^r$;

(1) If $r$ is even, the coefficient of $x^r$ in the second series is $4{(-1)}^{\frac{r}{2}}$; therefore in the expansion the coefficient of $x^r$ is $3+4{(-1)}^{\frac{r}{2}}$

So, the general term is $(3+4{(-1)}^{\frac{r}{2}})x^r$
(2) If $r$ is odd, the coefficient of $x^r$ in the second series is $-3{(-1)}^{\frac{r-1}{2}}$; therefore in the expansion the coefficient of $x^r$ is $3{(-1)}^{\frac{r+1}{2}}-3$

Thus, the general term is $(3{(-1)}^{\frac{r+1}{2}}-3)x^r$.