How can you prove n= $6k +2$ when the $x^2$ and $x^3$ coefficient of $(2k +x)^n$ is the same?

Question

I've tried working out the coefficients for each

$$\binom{n}{2}(2k)^{n-2}=\binom{n}{3}(2k)^{n-3}$$

But I'm not sure if that's helpful in any way? And I don't know what to do next?

Answer 1

Expand the binomials and cancel the $2k$'s: $$\frac{n(n-1)}2(2k)=\frac{n(n-1)(n-2)}6$$ We can cancel $n(n-1)$ too, and multiply by $6$: $$3(2k)=n-2$$ $$n=6k+2$$

Answer 2

Use the formula ${n\choose 2}={n(n-1)\over 2}$ and ${n\choose 3}={n(n-1)(n-2)\over 6}$.

Plug into your result you get $${n(n-1)\over 2}\times {(2k)}^{n-2}={n(n-1)(n-2)\over 6}\times{(2k)}^{n-3}$$

Cancel out both side you get $6k=n-2$

Answer 3

Yes you're on the right track, we just need to simplify the equation.

First divide out by $(2k)^{n-3}$ and use the formula for binomial coefficients to get $$ \frac12 n(n-1) \times 2k = \frac16 n(n-1)(n-2)$$ $$ k = \frac16 (n-2)$$ $$ n = 6k+2$$