Yes, that is correct. I would rather write 2^k+1 -2 as 2^(k+1) - 2 instead.
can anyone tell me if this seems correct? been working on this for a while now. Just want to make sure. I struggle with these quite a bit.
1.) 2 + 2^2 + 2^3 + 2^4 + ... + 2^n = 2^n+1 -2
Let n=1
2 + 2^2 + 2^3 + 2^4 + ... + 2^n = 2^1 = 2
and
2^n+1 -2 = 2^1+1 - 2 = 2^2 -2 = 4 - 2 = 2
assume n=k
2 + 2^2 + 2^3 + 2^4 + ... + 2^k = 2^k+1 -2
Let n=k+1
2 + 2^2 + 2^3 + 2^4 + ... + 2^k = 2^k+1 -2
= [2 + 2^2 + 2^3 + 2^4 + ... + 2^k] + 2^k+1
= [2^k+1 -2] + 2^k+1
= 2 x 2^k+1 - 2
= 2^1 x 2^k+1 - 2
= 2^k+1+1 - 2
= 2^(k+1)+1 - 2
Is the point of this part to show that 2^1 +...+ 2^k = 2^(k+1)-2 by showing that they both = 2 when n=1?Let n=1
2 + 2^2 + 2^3 + 2^4 + ... + 2^n = 2^1 = 2
and
2^n+1 -2 = 2^1+1 - 2 = 2^2 -2 = 4 - 2 = 2
If so then I can see how the case where n=k and n=k+1 works, but just because the equality holds when n=1 that doesn't necessarily means it must hold when n=k or n=k+1?
I'm new to proofs so that's why I ask, I'm not saying it's wrong, but that bit doesn't make sense to me.
He is showing two things:
1) This holds for n = 1
2) If this holds for n = k, then n = k + 1.
He is not using the n = 1 case to prove the n = k or n = k + 1 cases. He is assuming that this holds for some n = k. Then uses this assumption to prove the case where n = k + 1.
Note that when these two conditions are satisfied, then it is true for all n => 1. That is because by (1), we see this holds for n = 1. Then by (2) since this holds for n = 1, then it holds for n = 2. Then by (2) this holds for n = 3. Then by (2) again, this holds for n = 4 and so on.