# Prove

• Nov 22nd 2008, 10:15 AM
captainjapan
Prove
1.
1^3 + 2^3 + ... + n3 = (n(n+1)/2 )^2 for any positive integer n.

2.
3^n < n! for all n > 6 with n ∈ N.

mathematical induction to prove these?
• Nov 22nd 2008, 10:32 AM
o_O
Yes that's the way to go. Without going through all the formalities ...

#1: Assume it holds for $k$. We want to show that it holds for $k+1$ as well, that is, we want to show that $1^3 + 2^3 + \cdots + k^3 + (k+1)^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2$
So for your inductive step: ${\color{red}1^3 + 2^3 + \cdots + k^3} + (k+1)^3 = {\color{red}\left(\frac{k(k+1)}{2}\right)^2} + (k+1)^3 = \cdots$

#2: Again, assume it holds for $k$ for all $k > 6$, so we're assuming that $3^k < k!$. We then want to show that it is true for $k+1$ as well, that is, $3^{k+1} < (k+1)!$

Note that: $3^{k+1} = 3^k{\color{blue}3^1} < 3^k{\color{blue}(k+1)} < \cdots$

We can assume the blue because we're given that the statement is true for values strictly greater than 6. So obviously, $3 < 6 < k+1$. Hopefully you'll see the conclusion directly after using the inductive step.