I'm probably just being foolish, but I'm having trouble proving it true for n+1, based on the assumption that it's true for n.
Thanks for your help.
Hi feyomi,
P(k)
$\displaystyle k^3\ <\ 3^k$
P(k+1)
$\displaystyle (k+1)^3\ <\ 3^{k+1}$
Proof
Examine P(k+1) to see if P(k) being true will cause P(k+1) to be true
$\displaystyle (k+1)^3=k^3+3k^2+3k+1$
$\displaystyle 3^{k+1}=(3)3^k=3^k+3^k+3^k$
Hence, if $\displaystyle k^3\ <\ 3^k$
we ask if $\displaystyle 3k^2+3k+1\ <\ 3^k+3^k$
If $\displaystyle k\ \ge\ 4,\ 3k^2\ <\ 4k^2\ \Rightarrow\ 3k^2\ <\ k^3$
hence we ask if $\displaystyle 3k+1\ <\ 3^k$
$\displaystyle 3k+1\ <\ (k)k^2\ ?$
$\displaystyle 3k\ <\ k^2\ for\ k\ \ge\ 4$
hence $\displaystyle 3k+1\ <\ k^3,\ for\ k\ \ge\ 4$