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# Help GCSE maths question watch

1. Prove that (4n+1)^2 - (4n-1)^2 is always a multiple of 8, for all positive integer values of n.
2. (Original post by azo)
Prove that (4n+1)^2 - (4n-1)^2 is always a multiple of 8, for all positive integer values of n.
What have you tried
3. I tried multiplying out both brackets but after simplifying I was left with just 2 . I'm guessing I need to factorise it somehow to get 8 on the outside of the brackets, but I'm not sure how
4. (Original post by azo)
I tried multiplying out both brackets but after simplifying I was left with just 2 . I'm guessing I need to factorise it somehow to get 8 on the outside of the brackets, but I'm not sure how
You have a square minus a square. What does that suggest you do?

Multiplying out should also work. Try posting your working.
5. (Original post by notnek)
You have a square minus a square. What does that suggest you do?

Multiplying out should also work. Try posting your working.
Can I just cancel out the squares, so I'm left with 4n+1+4n-1 which simplifies to 8n???
6. (Original post by azo)
I tried multiplying out both brackets but after simplifying I was left with just 2 . I'm guessing I need to factorise it somehow to get 8 on the outside of the brackets, but I'm not sure how
It can be done by expanding. Most likely errors are always in expanding the brackets properly or errors with subtracting minus values.

It can be done more efficiently if you think about how something like x^2 - 36 would factorise.

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7. (Original post by gdunne42)
It can be done by expanding. Most likely errors are always in expanding the brackets properly or errors with subtracting minus values.

It can be done more efficiently if you think about how something like x^2 - 36 would factorise.

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(4n+1)(4n+1)-(4n-1)(4n-1)=16n^2+8n+1-16n^2-8n+1=2

That would factorise to (x+6)(x-6), I think.
8. (Original post by AlphaNick)
(4n+1)^2 = 16n^2 + 8n + 1
(4n-1)^2 = 16n^2 - 8n + 1

If you subtract equation b from equation a you'll get 16n
Remember that subtracting a negative gives a positive...

From here it should be obvious.
Cheers fella
9. (Original post by azo)
(4n+1)(4n+1)-(4n-1)(4n-1)=16n^2+8n+1-16n^2-8n+1=2

That would factorise to (x+6)(x-6), I think.
As pointed out 16n^2+8n+1-[16n^2-8n+1]= 16n = 8 x 2n ie a multiple of 8 for any positive integer value of n

You can alternatively use the difference of two squares approach here because the problem is also a square - a square

a^2 - b^2 = (a+b)(a-b)

[(4n+1)+(4n-1)] [(4n+1)-(4n-1)] =

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Updated: January 20, 2014
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