Question

Find the coefficient of
$x^8$ in the expansion of ${(1-\frac{x^2}{3}+\frac{x^4}{9})}^{-2}$.

Answer 1

${(1-\frac{x^2}{3}+\frac{x^4}{9})}^{-2}\equiv {(1-\frac{y}{3}+\frac{y^2}{9})}^{-2}$

Therefore we need to find the coefficient of $y^4$

General term of expansion is $\frac{(-2)(-3)...(-2-p+1)}{b!c!}{(-\frac{1}{3})}^b{\left(\frac{1}{9}\right)}^c{y}^{b+2c}$, where $p=b+c$

We want to find the coefficient of $y^4$, therefore $b+2c=4$. This is possible for,

$p=2,b=0,c=2;p=3,b=2,c=1;p=4,b=4,c=0$

$\therefore$ the coefficient of $y^4=\frac{(-2)(-3)}{2!}{\left(\frac{1}{9}\right)}^2+\frac{(-2)(-3)(-4)}{2!}{(-\frac{1}{3})}^2\left(\frac{1}{9}\right)+\frac{(-2)(-3)(-4)(-5)}{4!}{(-\frac{1}{3})}^4$

$=\frac{3}{81}-\frac{12}{81}+\frac{5}{81}=-\frac{4}{81}$