I was looking over this problem and im pretty sure you use induction.. but what would your base case be and when subing (k+1) for n, what is the inital formula we want to achieve? I dont exactly know how to approach this problem.
I was looking over this problem and im pretty sure you use induction.. but what would your base case be and when subing (k+1) for n, what is the inital formula we want to achieve? I dont exactly know how to approach this problem.
Hello jess0517
We have to prove that is an integer for all .
First we factorise:
So we must prove that is a multiple of 6; in other words, that it is a multiple of 2 and of 3.
First, note that one of and is even. So is even.
Next, note that is either a multiple of 3, in which case there is nothing left to prove; or it leaves a remainder 1 or 2 when divided by 3.
Case I: leaves a remainder 1; in which case . Therefore
and is therefore a multiple of 3.
Case II: leaves a remainder 2; in which case . Therefore
and is therefore a multiple of 3.
This concludes the proof.
Grandad
P(k)
is an integer ?
P(k+1)
is also an integer ?
Try to prove that P(k) being true causes P(k+1) to be true.
Proof
If k is an integer, so is k+1 and (k+1)(k+1).
Hence, if P(k) is true, it causes P(k+1) to be true.
All that remains is to test the proposition P(k) for the first integer.