It says in the solution that the generating function is:
$j=n-2$
$$x^2 \sum_{j=0}^\infty(2j+5)2^jx^j=2x^2\sum_{j=0}^\infty(j+1)(2x)^j+3x^2\sum_{j=0}^\infty(2x)^j=\frac{2x^2}{(1-2x)^2}+\frac{3x^2}{1-2x}$$
I don't understand the these steps. Is there some kind of summation rule that has been applied? Is there any other solution that is easier to follow?
The second sum is just a geometric sequence.
The first sum might me calculated like this
$$2x^2 \sum_{j=0}^{\infty}(j+1)(2x)^j =x^2 \sum_{j=0}^{\infty}\left( (2x)^{j+1} \right)’= x^2 \left(\sum_{j=0}^{\infty}(2x)^{j+1}\right)’ =x^2 \left(\frac{1}{1-2x}\right)’$$