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STEP 1 maths. Stuck on a proof

So, I don't need any actual help illustrating the proof, rather I need help understanding why it needs to be constructed in such a way. The question is hard to type so below is a link to the paper. It is Question 6 in the pure section.

https://3110b3b48cdc993f8fb622042e2ffa4f00a024fe.googledrive.com/host/0B1ZiqBksUHNYMTZlRm1iQnJPTWs/Papers/2013%20STEP%201.pdf

So I've done the first part but am having trouble understanding why you have to consider n as odd and even integers when outlining a proof for the second part. I do't understand why m cannot be 0..giving the b0 and b1 terms..then m = 1, giving the b2 and b3 term and so on...

I'm getting really confused, so thanks for any help!
Since B_n is defined differently depending on whether n is odd or even, it is going to be difficult to prove anything about B_n without considering odd and even cases at some point. You can consider n=1, n=2, n=3... (rather than n = 1, 3, 5, ... and n = 2,4,6...) but at some point you're probably going to find yourself saying "Now if n is odd, then ..., otherwise ...."

It's hard to be more specific when you've not actually given any details about what you're trying to do to prove the second part.
Original post by Student10011
So, I don't need any actual help illustrating the proof, rather I need help understanding why it needs to be constructed in such a way. The question is hard to type so below is a link to the paper. It is Question 6 in the pure section.

https://3110b3b48cdc993f8fb622042e2ffa4f00a024fe.googledrive.com/host/0B1ZiqBksUHNYMTZlRm1iQnJPTWs/Papers/2013%20STEP%201.pdf

So I've done the first part but am having trouble understanding why you have to consider n as odd and even integers when outlining a proof for the second part. I do't understand why m cannot be 0..giving the b0 and b1 terms..then m = 1, giving the b2 and b3 term and so on...

I'm getting really confused, so thanks for any help!

It's probably just easier to do it for odd and then for even integers. I haven't done the question, but you propose using induction in the "standard" way (n -> n+1 -> n+2 …) which tends to be a bit harder if you're trying to prove properties which are defined differently for odd vs even entries in a sequence.
Original post by Student10011


So I've done the first part but am having trouble understanding why you have to consider n as odd and even integers when outlining a proof for the second part.


Because the definition of BiB_i is different depending on whether i is odd or even.


I do't understand why m cannot be 0..giving the b0 and b1 terms..then m = 1, giving the b2 and b3 term and so on.


Well BiB_i is certainly defined for i= 0 and 1.

However, if you're refering to the formula Bn+2−Bn+1=BnB_{n+2}-B_{n+1}=B_{n} then the proof of that uses the first part, which is only defined for 1≤n1\leq n.
No doubt it can be tweeked for the extra cases.

Edit: There were no replies when I started typing - lol.
(edited 9 years ago)
Reply 4
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Student10011
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Original post by DFranklin
Since B_n is defined differently depending on whether n is odd or even, it is going to be difficult to prove anything about B_n without considering odd and even cases at some point. You can consider n=1, n=2, n=3... (rather than n = 1, 3, 5, ... and n = 2,4,6...) but at some point you're probably going to find yourself saying "Now if n is odd, then ..., otherwise ...."

It's hard to be more specific when you've not actually given any details about what you're trying to do to prove the second part.


I'm still a little unsure what you mean. I can follow the logic of the proof, but I'm confused as to how n differing gives different terms. I have to prove the the i'th term + (i+1)'st term = (i+2)'nd term. When m = 0, we have an expression for b_0 and b_1. Similarly for m = 1, we have an expression for b_2 and b_3, and so on, so I don't understand why I can't just prove b_2m + b_2m+1 = b_2m+2. Sorry If I'm being confusing!
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james22
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Original post by Student10011
I'm still a little unsure what you mean. I can follow the logic of the proof, but I'm confused as to how n differing gives different terms. I have to prove the the i'th term + (i+1)'st term = (i+2)'nd term. When m = 0, we have an expression for b_0 and b_1. Similarly for m = 1, we have an expression for b_2 and b_3, and so on, so I don't understand why I can't just prove b_2m + b_2m+1 = b_2m+2. Sorry If I'm being confusing!


Because then you have ruled out half the cases, e.g. n=1 (any odd n has been ruled out in fact).
Original post by Student10011
I'm still a little unsure what you mean. I can follow the logic of the proof, but I'm confused as to how n differing gives different terms. I have to prove the the i'th term + (i+1)'st term = (i+2)'nd term. When m = 0, we have an expression for b_0 and b_1. Similarly for m = 1, we have an expression for b_2 and b_3, and so on, so I don't understand why I can't just prove b_2m + b_2m+1 = b_2m+2. Sorry If I'm being confusing!

Can you give us your attempt at a proof using this method?
Reply 7
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Student10011
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Original post by james22
Because then you have ruled out half the cases, e.g. n=1 (any odd n has been ruled out in fact).


Oh yes...now I see what you mean. Seems so obvious now :colondollar:. Thanks!
Reply 8
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Student10011
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Thanks for the help with the question guys, got there in the end! Cheers to all :biggrin:

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