Find a closed form of the generating function for $a(n) =3n^2+4n+5$, $n=0,1,2,....$

Question

Find a closed form of the generating function for $a(n)=3n^2+4n+5$ n=0,1,2,....

not sure how the constraint n=0,1,2,.... applies to this. Does this means that the coefficient must go up sequentially starting from 0?

Answer 1

Let $$f(z) = \sum_{n=0}^\infty a_nz^n = \sum_{n=0}^\infty (3n^2 + 4n + 5)z^n.$$ Note that we can write $$ 3n^2 + 4n + 5 = 3n(n+1) + n + 5.$$ Hence $$\begin{align*} f(z) &= \sum_{n=0}^\infty(3n(n+1) + n + 5)z^n\\ &= 3\sum_{n=0}^\infty n(n+1)z^n + \sum_{n=0}^\infty nz^n + 5\sum_{n=0}^\infty z^n\\ &= 3z\sum_{n=0}^\infty (n+1)(n+2)z^n + z\sum_{n=0}^\infty (n+1)z^n + 5\sum_{n=0}^\infty z^n\\ &= \frac{6z}{(1-z)^3} + \frac z{(1-z)^2} + \frac5{1-z}\\ &= \frac{4z^2 - 3z + 5}{(1-z)^3}. \end{align*}$$

Answer 2

If you are allowed to use standard/well-known generating functions such as: $n^2=\frac{x (x+1)}{(1-x)^3};~n=\frac{x}{(1-x)^2};~1=\frac{1}{1-x}$, then you can simply calculate $f(x)$ as $$f(x)=3\left[\frac{x (x+1)}{(1-x)^3}\right]+4\left[\frac{x}{(1-x)^2}\right]+5\left[\frac{1}{1-x}\right]=\frac{(3-4 x) x-5}{(x-1)^3}.$$