Question

Expand $\frac{2+x^2}{1+x-x^2}$ in a series of ascending powers of $x$ as far as the term involving $x^5$.

Answer 1

Let $\frac{2+x^2}{1+x-x^2}=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots$,
where $a_0,a_1,a_2,a_3,\dots$ are constants whose values are to be determined; then
$2+x^2=(1+x-x^2)(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots )$.
In this equation we may equate the coefficients of like powers of $x$ on each side. On the right-hand side the coefficient of $x^n$ is $a_n+a_{n-1}-a_{n-2}$, and therefore, since $x^2$ is the highest power of $x$ on the left, for all values of $n>2$ we have
$a_n+a_{n-1}-a_{n-2}=0$;
this will suffice to find the successive coefficients after the first three have been obtained. To determine these we have the equations
$a_0=2$, $a_1+a_0=0$, $a_2+a_1-a_0=1$;
whence $a_0=2$, $a_1=-2$, $a_2=5$.
Also $a_3+a_2-a_1=0$, whence $a_3=-7$;
$a_4+a_3-a_2=0$, whence $a_4=12$;
and $a_5+a_4-a_3=0$, whence $a_5=-19$;
Thus $\frac{2+x^2}{1+x-x^2}=2-2x+5x^2-7x^3+12x^4-19x^5+\cdots$