I'm doing a coursework assignment and find myself rather stuck. I thought I understood back-substitution as a method for solving recurrences but am not finding my working to be getting me anywhere. My current question is just "What am I doing wrong in my working?". Here is my current working.
Thank you so much in advance.
The problem is $f(1)=2, f(2)=6, f(n)=f(n-2)+3n$
My working is:
$f(n)=f(n-2)+3n$ $=f(n-4)+3(n-2)+3n=f(n-4)+6n-6$ $=f(n-6)+3(n-4)+6n-6=f(n-6)+9n-18$ $=f(n-8)+3(n-6)+9n-18=f(n-8)+12n-36$ $...=f(n-2k)+3kn-3k(k-1)$
then put $k= {n-1 \over 2}$: $f\left(n-2{n-1\over 2}\right)+3\left({n-1 \over 2}\right)n-3\left({n-1 \over 2}\right)\left({n-1 \over 2}-1\right)$ $=...={1 \over 4}\left(9n^2-18n+17\right)$
But this does not check for $n=2$ or $n>3$
And yes, I have no choice but to use this method or similar to "guess" the form then prove it by induction.