Derive a recursive form of the function f(n) = 2n(n-6).

Question

The Function f: N -> Z is defined by f(n) = 2n(n-6) , for each integer n >= 0. Derive a recursive form of this function f. Please help :[

Answer 1

We have $f(n)=2n(n-6)=2n^2-12n$. By computing

\begin{equation} f(n+1) = 2(n+1)(n+1-6)=2n^2-8n-10, \end{equation}

we find that

\begin{equation} f(n+1)-f(n) = 4n - 10 = 4(n+1)-14, \end{equation}

and thus that \begin{equation} f(n+1) = f(n) + 4(n+1) - 10. \end{equation}

So for $n \geq 1$ we may write

\begin{equation} f(n) = 4n - 14 + f(n-1). \end{equation}

Answer 2

A two-term version:

$\begin{align} f(n)&=2n(n-6)\\ &=2n^2-12n \\ f(n-1) &= 2(n-1)(n-1-6) = (2n-2)(n-7) \\ &= 2n^2 - 16n+14 \\ f(n-2) &= 2(n-2)(n-2-6) = (2n-4)(n-8) \\ &= 2n^2 - 20n+32 \\ 2f(n-1)-f(n-2) &= 4n^2 - 32n +28 - ( 2n^2 - 20n+32) \\ &= 2n^2-12n -4 \\ &=f(n)-4 \\[0.5em] \hline \therefore f(n) &= 2f(n-1)-f(n-2)+4 \end{align}$