<s>[INST] 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 $f(n)$?
 [/INST] 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  
$f(x) - 1 = 3*(x-1)x \implies f(x) = 3x^2 - 3x + 1$ 
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