Posted: Nov 16, 2019 9:00 pm
The first thing I did was let z = (4k+1)1/2 as well. I also took the multilayering out to make the equation look like:
The binomial expansion for that sees the even powers of z cancel and leaves an expression:
With number of terms defined by the highest power being one less than or equal to n.
If you let x = z2 then the square roots go away nicely as well. The only outstanding problem is the 2n-1 in the denominator, and I felt like that must be a clue as to the association with Pascal's triangle, since the lines add up to the powers of 2, and we always have a list of combinations which constitute exactly half a row of Pascal's triangle, but with entries multiplied by various powers of x. Obviously those powers of x always produce binomial expansions with a 1 in them, which we could pull out, giving a full list of half of line n of Pascal's triangle, which is thus divisible by 2n-1, but what's left is looking like a mess.
So I couldn't see how to make an induction step in k from that. I'm not sure if this is even a step in the right direction.
The binomial expansion for that sees the even powers of z cancel and leaves an expression:
With number of terms defined by the highest power being one less than or equal to n.
If you let x = z2 then the square roots go away nicely as well. The only outstanding problem is the 2n-1 in the denominator, and I felt like that must be a clue as to the association with Pascal's triangle, since the lines add up to the powers of 2, and we always have a list of combinations which constitute exactly half a row of Pascal's triangle, but with entries multiplied by various powers of x. Obviously those powers of x always produce binomial expansions with a 1 in them, which we could pull out, giving a full list of half of line n of Pascal's triangle, which is thus divisible by 2n-1, but what's left is looking like a mess.
So I couldn't see how to make an induction step in k from that. I'm not sure if this is even a step in the right direction.