1. ## Induction proof...I'm stuck!

Hopefully someone can help me with this..I'm nearly done.

I'm in the induction step of the proof & here's what I have:

"So suppose that P(n) holds for n an integer and n>=5, we have that
2^n > n^2.
We wish to show that P(n+1) holds, that is that
2^(n+1) > (n+1)^2. This can be rewritten as
2(2^n) > n^2 +2n +1.
So, by our induction hypothesis and our Lemma (stated & proved earlier before this proof) we have
2(2^n) > 2^n > n^2 > 2n+1."

....This is where I'm stuck...is there a property that allows me to then just state "Thus 2(2^n) > n^2 +2n +1"?
Thanx for any feedback you provide!

2. $2^{n+1}=2^n+2^n>n^2+2n+1\rightarrow 2^n>2n+1$ using the hypothesis. This should be proved for $n\ge 5$: Suppose that $2^n>2n+1$; we show that $2^{n+1}>2(n+1)+1$. This is equvivalent to: $2^n+2^n>2n+3\rightarrow 2^n>2$ using the hypothesis. This is true even for $n\ge 1$. QED

3. Are you using the fact that 2^(n+1) = 2^n + 2^n ? I started to use this in my proof then realized that although this is true for a base 2 it is not true for a base > 2 (i.e. 3^(n+1) does not = 3^n + 3^n...
3^(n+1)=3(3^n).

4. You know (used it) that $2^{n+1}=2\cdot 2^n$. If you imagine that $2^n$ is an apple then $2\cdot 2^n$ is two apples so it apple+apple. Hope you understand it.
And yes of course $3^n+3^n=2\cdot 3^n$.

5. Originally Posted by james_bond
You know (used it) that $2^{n+1}=2\cdot 2^n$. If you imagine that $2^n$ is an apple then $2\cdot 2^n$ is two apples so it apple+apple. Hope you understand it.
And yes of course $3^n+3^n=2\cdot 3^n$.
Thanks James Bond!