Alternative for calculating the nth of quadratic sequence Given the quadratic sequence
$$f(n)=1, 7, 19, 37, \cdots$$
To calculate the $f(n)$ for $n\ge1$. $$f(n)=an^2+bn+c$$
We start with the general quadratic function, then sub in for $n:=1,2$ and $3$
$$f(1)=a+b+c$$
$$f(2)=4a+2b+c$$
$$f(3)=9a+3b+c$$
Now solve the simultaneous equations
$$a+b+c=1\tag1$$
$$4a+2b+c=7\tag2$$
$$9a+3b+c=19\tag3$$
$(2)-(1)$ and $(3)-(2)$
$$3a+b=6\tag4$$
$$5a+b=12\tag5$$
$(5)-(4)$
$$a=3$$
$$b=-3$$
$$c=1$$
$$f(n)=3n^2-3n+1$$
This method is very long. Is there another easy of calculating the $f(n)$?

A better explained version of @trancelocation's answer:
$$\begin{array}{}
1 & & 7 & & 19 & & 37  \\ 
  & 6 & & 12&  &18 &   \\ 
    & & 6 & & 6 & &   \\
\end{array}$$
The second row has equation $6n$. Therefore, $\big(a(n+1)^2+b(n+1)+c \big) - \big(an^2+bn+c \big)$ $ = 6n$, and so:
$$\big(a(n^2+2n+1)+b(n+1)+c \big) - \big(an^2+bn+c \big) = 6n$$
$$(2n+1)a + b = 6n$$
$a=3$ gives $6n+3$, and so $b=-3$. 'Plugging' $n=1$ into $3n^2-3n$ gives $0$, $c=1$, which gives $3n^2-3n+1$.
