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?

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- Nov 22nd 2008, 09:15 AMcaptainjapanProve
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, 09:32 AMo_O
Yes that's the way to go. Without going through all the formalities ...

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

So for your inductive step: $\displaystyle {\color{red}1^3 + 2^3 + \cdots + k^3} + (k+1)^3 = {\color{red}\left(\frac{k(k+1)}{2}\right)^2} + (k+1)^3 = \cdots$

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

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

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