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