Alternative for calculating the nth of quadratic sequence

Yes!

You know $f(1)= 1 \implies f(1) - 1 = 0 \implies f(x) - 1$ has a root at x=1

Now,

$f(x) - 1 = a(x-1)(x-b)$

Put $x = 2$, $a(2-b) = 6$

Put x = 3, $a*2*(3-b) = 18$

Divide both equation, we get $b = 0 $ and a = 3

$f(x) - 1 = 3*(x-1)x \implies f(x) = 3x^2 - 3x + 1$

Another standard way is to calculate a difference scheme and then to work backwards: $$\begin{matrix} 0 & & 1 & & 2 & & 3 & & 4 \\ & & 1 && 7 && 19 && 37 \\ && &6& & 12 && 18 & \\ &&&& 6 && 6 && \\ \end{matrix} \Rightarrow \begin{matrix} & 0 & & 1 & & 2 & & 3 & & 4 \\ \color{blue}{c}= &\color{blue}{1} & & 1 && 7 && 19 && 37 \\ \color{blue}{a+b}= & &\color{blue}{0}&&6& & 12 && 18 & \\ \color{blue}{2a}= &&& \color{blue}{6}&& 6 && 6 && \\ \end{matrix}$$ $$\Rightarrow a = 3, \; b= -3, \; c = 1 \Rightarrow f(n) = 3n^2-3n+1$$

Take a better basis. Namely, $\{(n-1)(n-2),(n-1)(n-3),(n-2)(n-3)\}$. If $$f(n) = \alpha(n-1)(n-2) + \beta(n-1)(n-3) + \gamma(n-2)(n-3),$$ then: $$f(1) = 0 + 0 + \gamma(1-2)(1-3),$$ $$f(2) = 0 + \beta(2-1)(2-3) + 0,$$ $$f(3) = \alpha(3-1)(3-2) + 0 + 0.$$