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FP1 Proof by Induction

Okay, so the question is 'prove by induction..' And I managed to do the whole proof except I'm stuck on the final part where you have to prove that n=k+1 should give you (k+1)((4k + 1)+1)((2k+1)+1)/3! Basically I'm stuck on the 'play around with algebra part' I tried factoring out (2k+1) (as shown in the zoomed up picture) but I felt like I wasn't really going anywhere! Is the any method/train if thought I should have when 'playing around with the algebra' because, in all honestly, I could be here forever trying to get where I need to be

Original post by creativebuzz

Not sure about anyone else but I'm finding it difficult to see your working / where you've done the 'assume that it holds for n=k' bit, and then the bit where you're trying k+1.

Reply 2

creativebuzz

Original post by SeanFM

Not sure about anyone else but I'm finding it difficult to see your working / where you've done the 'assume that it holds for n=k' bit, and then the bit where you're trying k+1.

Sorry, is this more clear? ( there,s a slight overlap with my working out for n=k+1 and n=1 because I did all the working it on one page)

Original post by creativebuzz

Sorry, is this more clear? ( there,s a slight overlap with my working out for n=k+1 and n=1 because I did all the working it on one page)

Yes, this is clearer. If the result is true, all terms on the LHS must factorise by (k+1). To check this, factorise out (1/3)(2k+1) (common term) from two of the terms and expand and then factorise what's inside of the brackets. Then factorise everything by (k+1) and carry on.

(edited 9 years ago)

Reply 4

creativebuzz

Original post by morgan8002

I managed to get this far, how can I take this further?

image.jpg75.9KB

Reply 5

username1560589

Original post by creativebuzz

I managed to get this far, how can I take this further?

You have the right idea now, but factorised (2k+1)/3 out incorrectly and the k shouldn't be factorised.
The second line should read:

\frac{2k+1}{3}[3(2k+1)+k(4k+1)] + 2^2(k+1)^2 = \frac{k+1}{3}(2k+3)(4k+5)

Reply 6

creativebuzz

Original post by morgan8002

You have the right idea now, but factorised (2k+1)/3 out incorrectly and the k shouldn't be factorised.
The second line should read:

\frac{2k+1}{3}[3(2k+1)+k(4k+1)] + 2^2(k+1)^2 = \frac{k+1}{3}(2k+3)(4k+5)

Ah I see! My main question is how did you know that you should factor the k back in to make k(4k+1) and to split (4k + 4) into 2^2(k+1)! I understand the maths around it but I don't see how I would look at my working out and know that I need to factor this and that out straight away to get my answe (if that makes sense)! Basically, I'm trying to adopt the right train of thought..

Reply 7

username1560589

Original post by creativebuzz

The whole point of these few lines are to find a way to factorise (k+1) from the whole expression. So basically I'm looking for (k+1) everywhere.

For the last bracket (2k+2)^2 it is fairly easy to see that you can factorise (k+1) out twice, which gives 2^2 = 4.

Because the left and the right have a factor (k+1) and the other term has a factor (k+1), the last two terms added together must have a factor (k+1) for it to work, so I'm looking for a (k+1) term there. The only common factor between these terms is (2k+1)/3 (k is not a factor of (2k+1)^2, so k cannot be factorised out).