- Prove that $(n+1)(n+2)(n+3)$ is $O(n^3)$
Attempt at Solution
- $f(n) = (n+1)(n+2)(n+3)$
- $g(n) = n^3$
Show that there exists an $n_0$ and $C > 0$ such that $f(n) \le Cg(n)$ whenever $n > n_0$
$f(n) = n^3+6n^2+11n+6 = n^3(1 + 6/n + 11/n^2 + 6/n^3)$
$f(n) \le C*g(n)$ is
$n^3(1 + 6/n + 11/n^2 + 6/n^3) \le C*n^3$ is
$(1 + 6/n + 11/n^2 + 6/n^3) < C$
That's as far as I got. Should I plug in a value for n to find C? And then, would that value I plugged in for n be $n_0$?
Any help is appreciated.
Thank you in advance.