Answer
See below
Work Step by Step
Given: $\sum (k+1)(k+2)a_{k+1}x^{k}-\sum ka_{k-1}x^k=0\\
\sum x^n[ (k+1)(k+2)a_{k+1}-ka_{k-1}]=0$
Comparing $x^k$ on both sides:
$a_{k+1}=\frac{ka_{k-1}}{(k+1)(k+2)}$
It gives:
$a_3=\frac{1}{2.3}a_0\\
a_4=\frac{3}{4.5}a_2=\frac{1.3}{2.3.4.5}a_0\\
a_k=\frac{1.3.5...(2k-1)a_0}{(2k+1)!}$
and:
$a_3=\frac{1}{3.4}a_1\\
a_4=\frac{4}{5.6}a_3=\frac{4}{3.4.5.6}a_1\\
a_{2k+1}=2[\frac{2.4...(2k).a_0}{(2k+2)!}]\\
=\frac{2^{k+1}k!}{(2k+2)!}$
Therefore: $f(x)=a_0\sum \frac{1.3...(2k-1).a_0}{(n+1)!}x^{2k}+a_n \sum \frac{2^{k+1}k!}{(2k+2)!}x^{2k+1}$
