An algebra problem by Ojas Singh Malhi

Observe that the outputs of the polynomial follow the pattern $2x-1$ . But this isn't the answer to the question as it requires a $5th$ degree polynomial.
Therefore, we need a polynomial such that $x=1,2,3,4,5$ , gives $f(x)=2x-1$ .
Clearly, $f(x)=2009(x-1)(x-2)(x-3)(x-4)(x-5)+2x-1$ fulfils our requirements ( $5th$ degree polynomial with leading coefficient $2009$ ).
Now, $f(6)=2009*5*4*3*2*1+11$
$f(6)=241091$