1. ## Induction

2. Originally Posted by ff4930
Im confused at this problem, any help would be appreciated.

Let
$\displaystyle a_{1} = a$

$\displaystyle a_{n} = 2a_{n-1} + 1$

Prove by induction that
$\displaystyle a_{n} = 2^(n) - 1$

Can a be any integer?
if I put $\displaystyle a_{1} = 5$
then$\displaystyle a_{1} = 5 = 5$
if I let$\displaystyle n = 2$
then
$\displaystyle a_{2} = 2(5)+1 = 11 != 2^(n) - 1 = 2^2-1 = 3$

that's true, $\displaystyle a_1 = 5$ does not work for this. are you sure you copied the question correctly? have you left any information or conditions out? i think it works for $\displaystyle a_1 = 1$ though, then the proof would be easy, in fact, i think we can show that $\displaystyle a_1 = 1$ is the only plausible first term

3. This is an print out review exam problem, I just checked online and she has took this problem out. I had the older version. I guess my prof. saw the problem, sorry about that.

If you guys can help me on my second most frustrated Induction, it be great.

Use induction to prove that n! < n^n whenever n is a positive integer greater than 1.

Base case: n = 2
2! = 2 < 2^2 = 4 - checks!

Assume n! < n^n, n = k.

Prove that n! < n^n when n = k+1

4. Say that $\displaystyle k > 1\quad \& \quad k! < k^k$.
Look at this.
$\displaystyle \left( {k + 1} \right)! = \left( {k + 1} \right)\left( {k!} \right) < \left( {k + 1} \right)\left( {k^k } \right) < \left( {k + 1} \right)\left( {\left[ {k + 1} \right]^k } \right)$.

Can you finish?

5. Originally Posted by Plato
Say that $\displaystyle k > 1\quad \& \quad k! < k^k$.
Look at this.
$\displaystyle \left( {k + 1} \right)! = \left( {k + 1} \right)\left( {k!} \right) < \left( {k + 1} \right)\left( {k^k } \right) < \left( {k + 1} \right)\left( {\left[ {k + 1} \right]^k } \right)$.

Can you finish?
I'm sorry I'm just so bad in Induction. I can keep staring and I still wont see it.

Did you add (k+1) to both sides?
I lost you when you got up to $\displaystyle \left( {k + 1} \right)\left( {\left[ {k + 1} \right]^k } \right)$
Could you explain on how you got that please.

6. It is really so simple!
$\displaystyle k + 1 > k\quad \Rightarrow \quad k^k < \left( {k + 1} \right)^k$

7. wrong post sorry