Binomial Theorem expansion with algebra

[thomas49th]

In the binomial expansion [tex](2k+x)^{n}[/tex], where k is a constant and n is positive integer, the coefficient of x² is equal to the coefficient of x³

a) Prove that n = 6k + 2
b) Given also that [tex]k = .\frac{2}{3}[/tex], expand [tex](2k+x)^{n}[/tex] in ascending powers of x up to and including the term in x³, giving each coefficient as an exact fraction in its simplest form.

my shot at (a)

expand to get x² and x³:

[tex]\stackrel{n}{2}(2k)^{n-2}[/tex] + [tex]\stackrel{n}{3}(2k)^{n-3}[/tex]

subst n = 6k + 2 but i get into more expansion, which i don't think is really going anywhere

can someone help me out/guide me through (a) please?

Thankyou

 

[arildno]

So, do you agree that we must have:
[tex]\frac{n!}{2!(n-2)!}=\frac{n!}{3!(n-3)!2k}[/tex]

Clearly, this can be simplified to:
[tex]\frac{3!2k}{2!}=\frac{(n-2)!}{(n-3)!}[/tex]

Can you simplify this into your desired result?

 

[thomas49th]

i can get it [tex]\frac{3!2k}{2!}=\frac{(n-2)!}{(n-3)!}[/tex] down to 6k(n-3)=1 but that won't simplify to n = 6k + 2.

 

[arildno]

Now, you are muddling!
Which number is biggest: (n-2)! or (n-3)!

 

[thomas49th]

(n-2)(n-1) /
(n-3)(n-2)(n-1)

so (n-2)! is cancels

so leaves 1/(n-3)

right?

 

[arildno]

No.
We have: (n-2)!=(n-2)*(n-3)!