A level maths, binomial expansion!?

Notation: I'm using nCd to represent the number of possible combinations of n items taken d at a time. (I need that to represent coefficients in the expansion; the most customary notation doesn't work in Answers.)

We know the general formula for those is

nCd = n! /[(n-d)! d!]

The first few terms of the expansion of

(2k+x)^n will be

nC0 (2k)^n + nC1 (2k)^(n-1) x + nC2 (2k)^(n-2) x^2 + nC3 (2k)^(n-3) x^3 + ...

and we have the x^2 and x^3 terms, for which we know the coefficients are equal.

a) So

nC2 (2k)^(n-2) = nC3 (2k)^(n-3)

nC2 (2k) = nC3 [dividing both sides by (2k)^(n-3)]

2k = nC3 / nC2 [dividing both sides by nC2]

2k = {n! / [(n-3)! 3!]} / {n! / [(n-2)! 2!]} [using the formulas for nC3 and nC2]

2k = {n! / [(n-3)! 3!]} * {[(n-2)! 2! / n!} [dividing fractions: invert denominator & multiply]

2k = [(n-2)! 2!] / [(n-3)! 3!] [the n! terms cancel out]

2k = (n-2) / 3 [canceling out other duplicate terms in factorials]

6k = n - 2

6k + 2 = n

which is what we were to prove.

b) If k=2/3, then

n = 6k + 2 = (12/3) + 2 = 6

(2k + x)^n = [(4/3) + x]^6

= (4/3)^6 + 6 (4/3)^5 x + 15 (4/3)^4 x^2 + 20 (4/3)^3 x^3 + 15 (4/3)^2 x^4 + 6 (4/3) x^5 + x^6

= 4096/729 + (2048/81)x + (1280/27)x^2 + (1280/27)x^3 + (80/3)x^4 + 8x^5 + x^6

[Might as well do the WHOLE expansion.]

The first few terms of the binomial expansion of (2k + x)ⁿ are :-

(2k)ⁿ + n(2k)ⁿ⁻¹x + n(n - 1)/2! *(2k)ⁿ⁻²x² + n(n - 1)(n - 2)/3! * (2k)ⁿ⁻³x³ +...

a) As the coefficients of x² and x³ are the same then :-

n(n - 1)/2! *(2k)ⁿ⁻² = n(n - 1)(n - 2)/3! * (2k)ⁿ⁻³

2k =(n - 2)/3 : 6k = n - 2 : n = 6k + 2

b) As n = 6k + 2 then when k = ⅔, n = 6.....and using the earlier expansion gives :-

(2*⅔)^6 + 6*(2*⅔)^5.x + 6*5/2! *(2*⅔)^4.x² + 6*5*4/3! * (2*⅔)³.x³

= (4096/729) + (2048/81)x + (1280/27)x² + (1280/27)x³........which is what you would expect as the coefficients of x² and x³ are equal.

The coefficient of x^2 is nC2(2k)^(n-2)---------------(a million) The coefficient of x^3 is nC3(2k)^(n-3)----------------(2) (a million)=(2)=> [n!/(n-2)!2!](2k)^(n-2)= [n!/(n-3)!3!](2k)^(n-3)=> 2k/(n-2)=a million/3=> 6k=n-2=> n=6k+2