Answer
$$\sum^{\infty}_{n=0}((n+2)(n+1)a_{n+2} - n(n-1)a_n)x^n $$
Work Step by Step
1. Expand the expression:
$$\sum^{\infty}_{n=2} n(n-1)a_nx^{n-2} -x^2\sum^{\infty}_{n=2} n(n-1)a_nx^{n-2} $$
$$\sum^{\infty}_{n=2} n(n-1)a_nx^{n-2} - \sum^{\infty}_{n=2} n(n-1)a_nx^{n} $$
2. On the first one, substitute each 'n' with a 'n+2'
$$\sum^{\infty}_{n+2=2}(n+2)(n+2-1)a_{n+2}x^{n+2-2} $$ $$\sum^{\infty}_{n=0}(n+2)(n+1)a_{n+2}x^{n} $$
3. Write the second expression starting at n = 0:
$$ \sum^{\infty}_{n=0} n(n-1)a_nx^{n} = (0)(0-1)a_0x^0 + (1)(1-1)a_1x^1 + \sum^{\infty}_{n=2} n(n-1)a_nx^{n}$$
$$ \sum^{\infty}_{n=0} n(n-1)a_nx^{n} = \sum^{\infty}_{n=2} n(n-1)a_nx^{n}$$
4. Substituting:
$$\sum^{\infty}_{n=0}(n+2)(n+1)a_{n+2}x^{n} - \sum^{\infty}_{n=0} n(n-1)a_nx^{n} $$
$$\sum^{\infty}_{n=0}[(n+2)(n+1)a_{n+2}x^{n} - n(n-1)a_nx^{n}] $$
$$\sum^{\infty}_{n=0}[(n+2)(n+1)a_{n+2} - n(n-1)a_n]x^n $$
