Answer
$=a_0(1+\frac{2}{3}x+\frac{1}{12}x^2)$
Work Step by Step
Since $n =0 \rightarrow n(n+2)a_{n}x^{n}=0$:
We have: $\sum n(n+2)a_{n}x^{n}=\sum n(n+2)a_{n}x^n$
Obtain:
$\sum n(n+2)a_{n}x^{n}+\sum (n-3)a_{n-1}x^n\\
=\sum [n(n+2)a_n+(n-3)a_{n-1}]x^n$
Comparing $x^n$ on both sides:
$a_n=\frac{(3-n)a_{n-1}}{n(n+2)}$
It gives:
$a_1=\frac{2}{3}a_0\\
a_2=\frac{1}{8}a_1=\frac{1}{12}a_0\\
a_3=0\\
a_4=0$
We have: $f(x)=a_0+\frac{2}{3}xa_0+\frac{1}{12}x^2a_0\\
=a_0(1+\frac{2}{3}x+\frac{1}{12}x^2)$
