1. ## Mathematical induction

Plese help me with this:

Use mathematical induction to prove this (n+1)^n < (n)^n+1 for all n> or =3.

Is anyone has an idea? I find it so hard.

2. ## Re: Mathematical induction

Have you tried anything so far? What is the base step?

3. ## Re: Mathematical induction

This is the first step. Suppose n=k. Where k is equal or greatee than 3.

(K+1)^k < (k)^k+1
[K(1+1/k)]^k < k^k *k
K^k *(1+1/k)^k <k^k *k
(1+1/k)^k < k

Trivial case: if k=3
(1+1/3)^3 < 3
2.37< 3

Second step: suppose n= k is true. Then n= k+1 must be also true.
I have a problem with this step. I cant prove it. 😥😥😥😥

4. ## Re: Mathematical induction

Method by trying to "work backwards":

You could do the inductive step in one line if it were true that: $\displaystyle \frac{(n+2)^{n+1}}{(n+1)^n} < \frac{(n+1)^{n+2}}{n^{n+1}}$ (for $\displaystyle n \ge 3$).

Do you see why? Since everything in sight is positive, you'd have, for $\displaystyle n \ge 3$, that:

$\displaystyle 0 < (n+1)^n < n^{n+1}$ (your inductive assumption) and $\displaystyle 0 < \frac{(n+2)^{n+1}}{(n+1)^n} < \frac{(n+1)^{n+2}}{n^{n+1}}$ (this wonderful thing that gets the job done if it happens to be true)

and therefore $\displaystyle 0 < \left( (n+1)^n \right) \ \left( \frac{(n+2)^{n+1}}{(n+1)^n} \right) < \left( n^{n+1} \right) \ \left( \frac{(n+1)^{n+2}}{n^{n+1}} \right)$,

which would prove that $\displaystyle 0 < (n+2)^{n+1} < (n+1)^{n+2}$.

And that would prove the induction.

Do you see why I called that working backwards? I chose to consider $\displaystyle \frac{(n+2)^{n+1}}{(n+1)^n} < \frac{(n+1)^{n+2}}{n^{n+1}}$ because, if it were true, it would be exactly what would be needed to get the job done.

It turns out that $\displaystyle \frac{(n+2)^{n+1}}{(n+1)^n} < \frac{(n+1)^{n+2}}{n^{n+1}}$ is true (actually for all $\displaystyle n \ge 1$), and you can prove that directly with some algebra. Can you see how?

5. ## Re: Mathematical induction

Im sosory. I just dont get it on how you started the process. 😥

6. ## Re: Mathematical induction

An alternative approach is to examine the binomial expansion$$(n+1)^{n+2}=\sum_{k=0}^{n+2}{n +2\choose k}n^k$$
We can use the inductive assumption on each term and the identity $${m+1 \choose i} = {m \choose i-1} + {m \choose i}$$
This should lead to something similar (but greater than) the binomial expansion of $$(n+2)^{n+1} = \big((n+1)+1\big)^{n+1}=\sum_{j=0}^{n+1} {n+1 \choose j}(n+1)^j$$

7. ## Re: Mathematical induction

Hard to tell if you REALLY can't...or are too lazy