1. ## Induction Help

Im trying solve 2 questions using induction.

Problem 1.

n equals 2^n where n is length of bit string (For example if n = 1, then only 0 and 1 are possible outcomes).

Base Case: Assume n = 1, therefore 2^1 = 2. Which is correct as i described above.

Induction Case: I get alittle troubled by something now..
Show that P(n) holds then so does P(n+1).

n, (n+1) = 2^(n+1)
Now I rearrange the RHS correct?
= (2n + 2)
=2(n+1)

Can somebody help me out if im on the right track on this?

Problem 2.
Similar problem but the numbers of binary words of length at most n is equal to 2^(n+1) - 1.
Base Case: n = 1
So 2^(1+1)-1 = 3
Shouldnt it be 2?

I know that both of these are correct i just have to prove them, but my mind gets muddled as i described above and id love some clarification if possible.
(Actually it could be that i need to prove them false).

Thanks for giving me your generous help.

2. I have no idea what you think the problem actually says.
BUT $\displaystyle \left( {\forall n \in \mathbb{N}} \right)\left[ {{\color{red}n \ne 2^n} } \right]$.

Why don't you quote the problem exactly as it was given to you?

3. Sorry, whats clear to me isnt always clear to someone else i suppose!

Problem 1: I need to use induction to prove that the number of binary strings of length n equals 2^n.

Problem 2.Number of binary strings of length at most n equals 2^(n+1) -1