creativebuzz
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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
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Kevin De Bruyne
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(Original post by creativebuzz)
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
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.
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creativebuzz
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(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.
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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)
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username1560589
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(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.
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creativebuzz
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(Original post by morgan8002)
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.
I managed to get this far, how can I take this further?
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username1560589
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(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)
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creativebuzz
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(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..
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username1560589
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(Original post by creativebuzz)
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..
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).
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creativebuzz
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(Original post by morgan8002)
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).
Ah I think I get it now!

could you explain what they did in these last two lines, I literally have no idea where the 55 and 63 came from..

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username1560589
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(Original post by creativebuzz)
Ah I think I get it now!

could you explain what they did in these last two lines, I literally have no idea where the 55 and 63 came from..
64(2^{6k}) - 2^{6k} = 63(2^{6k}) (where the first 63 came from)
To get the 63 and -55, they split the 8 up (8 = 63-55). So 8(3^{2k-2} ) = 63(3^{2k-2}) - 55(3^{2k-2})
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