Binomial expansions - struggling...

Hi, need some guidance if you don't mind.

Question.
In the B.E of (2k+x)^n where k is a constant and n is a positive integer, the co-efficient of x^2 is equal to the co-efficient of x^3.

PROVE that n = 6k+2!!

My answer so far using (a+b)^n
so

n(n-1)/2 * 2k^n-2 * x^2 = n(n-1)(n-2)/6 * 2k^n-2 * x^3


the answer in the booklet says the next step is...
2k = (n-2)/3

HOW does it go from my step the next step is where Im struggling?

Thanks

Hi, need some guidance if you don't mind.

Question.
In the B.E of (2k+x)^n where k is a constant and n is a positive integer, the co-efficient of x^2 is equal to the co-efficient of x^3.

PROVE that n = 6k+2!!

My answer so far using (a+b)^n
so

n(n-1)/2 * 2k^n-2 * x^2 = n(n-1)(n-2)/6 * 2k^n-2 * x^3


the answer in the booklet says the next step is...
2k = (n-2)/3

HOW does it go from my step the next step is where Im struggling?

Thanks

There should be no x in your equation

Also the 2k will have different powers - the second one should have n-3

Then - look for things you can cancel - both sides have n and (n-1) for example

thanks both.
Yes it was a typo error n-3 not n-2..

ok so i got rid of n(n-1)

1/2*2k^n-2 = (n-2)/6 * 2k^n-3

then 1/2 * 2 (in the first term)

k^n-2 = (n-2)/6 * 2k^n-3

right so far?

thanks both.
Yes it was a typo error n-3 not n-2..

ok so i got rid of n(n-1)

1/2*2k^n-2 = (n-2)/6 * 2k^n-3

then 1/2 * 2 (in the first term)

k^n-2 = (n-2)/6 * 2k^n-3

right so far?

You seem to have lost a multiple of 2 in there - I think it's because you're not using brackets or Latex!

Remember that you have $(2k)^{n-2}$ on the left and $(2k)^{n-3}$ on the right when you first start. Then you can cancel powers of 2k.

from
$\dfrac{n(n-1)}{2}\dfrac{(2k)^n}{(2k)^2} = \dfrac{n(n-1)(n-2)}{6}\dfrac{(2k)^n}{(2k)^3}$

cancel
n
n-1
(2k)^n

multiply both sides by 6
multiply both sides by (2k)^3

Oh god I did not know i could do that!!! I just need to get my head around how you have set out the
(2k)^n-1

(2k)^n/(2k)^2 just so im on the right track, this is relating to dividing powers. e.g n-2 thats cool, glad I know...

yes ive done the question now and I've sussed it.

Thanks