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Help GCSE maths question

azo

Prove that (4n+1)^2 - (4n-1)^2 is always a multiple of 8, for all positive integer values of n.

Reply 1

TenOfThem

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

Reply 2

azo

I tried multiplying out both brackets but after simplifying I was left with just 2 :frown:

. I'm guessing I need to factorise it somehow to get 8 on the outside of the brackets, but I'm not sure how :frown:

Reply 3

Notnek

Original post by azo

I tried multiplying out both brackets but after simplifying I was left with just 2 :frown:

. I'm guessing I need to factorise it somehow to get 8 on the outside of the brackets, but I'm not sure how :frown:

You have a square minus a square. What does that suggest you do?

Multiplying out should also work. Try posting your working.

(edited 10 years ago)

Reply 4

azo

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???

Reply 5

gdunne42

Original post by azo

I tried multiplying out both brackets but after simplifying I was left with just 2 :frown:

. I'm guessing I need to factorise it somehow to get 8 on the outside of the brackets, but I'm not sure how :frown:

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.

Posted from TSR Mobile

(edited 10 years ago)

Reply 6

azo

Original post by gdunne42

(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.

(edited 10 years ago)

Reply 7

azo

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

Reply 8

gdunne42

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)] =

Posted from TSR Mobile

(edited 10 years ago)