Quote:
Originally Posted by BruceZ
Yes, I used k=3 as the base case, so all you can logically conclude from that is that it's true for all n >= 3. I could have used k=2 as the base case and proved it was true for n >= 2. Even n >=1 if you consider 1 to be a sum of 1 number.
Do you see why it's logically valid now? Do you understand that's all we did? You need to understand this proof to be able to do any others. This is the simplest one, and the rest are all essentially the same just with more complicated algebra than noticing k^2 + 2k + 1 = (k+1)^2. In fact even if you didn't already know that, it's clear that is what we need to be true to establish (2), so then you'd set about showing that it is true, say by multiplying out (k+1)*(k+1). So you always know what you need to show to get the induction to work. Do you understand how we used that to establish that (2) is true?
Yes. I see it.
Here's an example that shouldn't be hard, but that I don't get anywhere with:
given a_1=1 and a_(n+1)=3-(1/a_ n)
I have an intuition that a_n < 3 for all n <=1
We know it's true for n=1
We assume it is true that a_k < 3
and need to show that the following is true
a_(k+1) < 3
My initial strategy is to manipulate the statement from the inductive hypothesis, while maintaining the equality, in order to derive the statement we want to prove is true.
a_k < 3
1/a_k > 1/3
3 - 1/a_k > 3 - 1/3
So at this point I've derived the statement I wanted to, but in doing so I wasn't able to prove what I thought was true.