Need help with some proofs!?

Question

I started 3 of these and I'm not sure that I'm on the right track.

1. Prove and extend or disprove and salvage:
For every natural number n, 21 + 22 + ... + 2n = 2n+1

2. Prove and extend or disprove and salvage:
For every natural number n > 1, either n is a prime, n is a perfect square, or n | (n−1)!

3. Prove and extend or disprove and salvage:
Let A and B be integers. A + B is even if and only if AB is odd.

Answer 1

1) I think you should rewrite this and didn't think you meant that.
I think you meant 2^n. In any case you want to mention what you mean by putting it in your details.

3)

This is false.
Counterexample: Let A = 2 and B = 4. Then
A + B = 2 + 4 = 6, but

AB = (2)(4) = 8, which is NOT odd.

Answer 2

(1) See, Chris, you do notice that 21,22,23,.........,2n forms an A.P., don't you?
The first term of this A.P. is a=21.
The second term is=22
So, common difference of this A.P. is d= 22-21 = 1
Let 2n be the mth term of this A.P.
Then, 2n=a+(m-1)d [Since, this is the formula for the general term (here, its the last term) of an A.P.]
So, 2n = 21 + (m-1)*1
i.e., 2n=21+m-1
i.e., m=2n-20-------------------->(i)
Now, the last term, which here is the mth term, is=2n
i.e., a+(m-1)d = 2n
i.e., 21+(2n-20-1)d=2n [Using (i)]
i.e., 21+2nd-21d=2n
i.e., 2n(d-1)=21(d-1)
i.e., 2n=21
i.e., n=10.5
But we know that a 'term' of an A.P. cannot be a fraction. For example, we have the 5th term of an A.P., we have 13th term of an A.P., but we can never have 7.8th term of an A.P. Know what I mean? So, what the fractional value of n implies is that the given A.P. cannot have its last term in the form of 2n. This fact can also be known from the fact that the first term is 21 & we obtained the last term to be 2n=21, which implies that the A.P. has only 1 term, & hence we can't find a sum of its terms as it has no other term. So, this has been disproved.

You can prove the other two using the Principle of Mathematical Induction. They are very easy & I believe you can do them on your own. So, I've not solved them for you. You can mail me if you are unable to solve them on your own.