Finding the value of $f(6)$ when $f(x)$ of degree $5$ with leading coefficient Problem : Suppose $f(x)$ is a polynomial of degree $5$, and with leading coefficient $2009$. If further that $f(1) =1; f(2)=3, f(3)=5, f(4)=7, f(5)=9$. What is the value of $f(6)?$ 
My work : 
Let $f(x) = ax^5+bx^4+cx^3+dx^2+ex+f$ 
Now 
$$\begin{align} f(1) &= 2009 \cdot1 +b +c+d+e+f = 1 . . . .(i)\\
f(2) &= 2009\cdot(2)^5+b(2)^4+c(2)^3+d(2)^2+2e+f....(ii)\\
f(3)& = 2009\cdot(3)^5+b(3)^4+c(3)^3+d(3)^2+3e+f....(iii)\\
f(4) &= 2009\cdot(4)^5+b(4)^4+c(4)^3+d(4)^2+4e+f....(iv)\\
f(5) &= 2009\cdot(5)^5+b(5)^4+c(5)^3+d(5)^2+5e+f....(v)\end{align}$$
Can we do this way by solving these five equations or is their any other easier way of doing  this.. Please suggest thanks a lot..

Consider the polynomial $h(x) = f(x)-(2x-1)$ which is also of degree $5$. You are given that $1,2,3,4,5$ are all the $5$ roots of $h$, and that it's leading coefficient is $2009$. Therefore,
$$h(x)=2009(x-1)(x-2)(x-3)(x-4)(x-5)$$
Now putting in $x=6$, $h(6)=2009*5*4*3*2*1=241080$.
This gives $f(6)=h(6)+(2\times 6-1)=241080+11=241091$
