C2 Binomial expansion mixed exercise

I was thinking about those who self teach.

I seem to have answered this question in another thread you've resurrected from a couple of years ago - you just need to divide both sides by $(2k)^{n-3}$. This shouldn't be anything new to you - it's just GCSE-level manipulation of indices

What are some good textbooks for C1 and C2 etc then? Or is it just a case of using your common sense to fill in the gaps of the textbook?

Okay so tell me where im going wrong then?

Okay so tell me where im going wrong then?

Fair Enough

The OP has a teacher though

I can barely read that, but I think if you simplify your fraction and multiply by 3 you've virtually got the equation you're trying to prove

I still say that I think some teachers may be right not to expose certain levels of mathematical understanding on some pupils.
It sounds controversial but often teaching less able pupils (which is in no way a reference to the OP ) often can lead to just algorithms, plugging and chugging and keeping things very basic just to ensure they access some of the content.
Dare I say it but I believe there will be a day not so far ahead where I teach differentiation by first principles outside of teaching hours.
I am not saying these methods are right but if you have pupils returning papers with less than 15/20 out of 75 in March/April do you go down the 'knowledge' route?

Yeah sorry about that couldnt find my camara so had to use phone, right okay, but if I have
n-2 x 6k = 0
can the 6k just be put onto the other side? so i can get to
n=6k+2

I know what you mean

You should read my tirade on the "do you get taught differentiation properly" thread

When you divided by 6n(n-2) you have 0 on the left hand side - this is incorrect

$\dfrac{a}{a} \not= 0$

"I jus wanna no wot dat dy/dx thing = "

Okay, so what is left on the left hand side?

Well, what is $\dfrac{a}{a}$?

Haha oh dear, um is it 1

Differentiate = Decrease

Integrate = Increase

Sorry to keep asking, but I've got to question 13 part b,
Q13 is:
a. Given that (2+x)⁵ + (2-x)⁵ = A + Bx² + Cx⁴
find the values of constants of A, B and C
Ive done this bit and got 64, 160 and 20, checked and they are correct
b. Using the substitution y=x² and your answers to part a, solve
(2+x)⁵+(2-x)⁵=349
so I've done
64+160x²+20x⁴=349
so 20x⁴+160x²-285=0
let y=x²
so 4y²+32y-57=0
but if i solve this in the quadratic function i get y=3/2 or -19/2 which gives weird answers for x and the answer in the book is x= +or- 3/2
Can you see where im going wrong?

Thanks

Cannot see any error

Okay thank you for checking, I appreciate your time, but if its correct how do I get to x=+or- 3/2?