Tony @ Tue Dec 10, 2013 2:45 am wrote:
You need to actually solve for the value of f(n-1). But you already know what it is!
If f(n) = n^2 + 2m * f(n-1)
then f(n-1) = (n-1)^2 + 2m * f(n-2)
which you can put back into the original equation.
f(n) = n^2 + 2m * ((n-1)^2 + 2m * f(n-2))
etc. Keep going until you remove f() from the right side of the equation, then simplify.
not sure how to "keep going and simplify". The third equation would be:
f(n) = n^2+2m[(n-1)^2+2m(n-1)^2+4m^2f(n-3)], .... what exactly would be the i'th case? I don't see a pattern here.. and why is this equation so tedious? Is this the right way to do a multiple choice question worth 1 mark on a 100 mark 3 hour final exam? Because I would probably get 0% on the final then haha