If $(6-12x+12x^2)^n = \displaystyle \sum_{r=0}^{2n} T_r x^r$, prove $ T_r = (-2)^r 3^n \bigg[\binom{2n}{r} + \binom{2n-2}{r}\binom{n}{1} + \binom{2n-4}{r}\binom{n}{2} + \dots \bigg]$
I tried by writing $$(6−12x+12x^2)^n= 6^n(1-2x+2x^2)^n = 6^n\displaystyle \sum_{a,b,c}^{a+b+c = n} \frac{n!}{a!b!c!}(-2x)^b(2x^2)^c = 6^n \sum_{a,b,c}^{a+b+c = n} \frac{n!}{a!b!c!}(-1)^b 2^{b+c} x^{b+2c}$$ which gives two conditions, $a+b+c = n$ and $b+2c =r$
So the (a,b,c)'s possible here are (n-r,r,0) ; (n-r+1,r-2,1) ; (n-r+2,r-4,2) ... (0,0,n)
So substituting those a,b,c's I get the coefficient of $x^r$ as $$6^n \bigg[\frac{n!}{(n-r)!r!0!}(-1)^r 2^{r} + \frac{n!}{(n-r+1)!(r-2)!1!}(-1)^{r-2} 2^{r-1} + \frac{n!}{(n-r+2)!(r-4)!2!}(-1)^{r-4} 2^{r-2} + \cdots \bigg] =$$ $$\displaystyle 6^n(-1)^r2^{r} \bigg[\frac{n!}{(n-r)!r!0!} + \frac{n!}{(n-r+1)!(r-2)!1!} 2^{-1} + \frac{n!}{(n-r+2)!(r-4)!2!} 2^{-2} + \cdots \bigg]$$ $ = \displaystyle (-2)^r 3^n \bigg[\frac{n!}{(n-r)!r!0!}2^n + \frac{n!}{(n-r+1)!(r-2)!1!} 2^{n-1} + \frac{n!}{(n-r+2)!(r-4)!2!} 2^{n-2} + \cdots \bigg]$ I guess I have to prove that $$\displaystyle \bigg[\frac{n!}{(n-r)!r!0!}2^n + \frac{n!}{(n-r+1)!(r-2)!1!} 2^{n-1} + \frac{n!}{(n-r+2)!(r-4)!2!} 2^{n-2} + \cdots \bigg] = \bigg[\binom{2n}{r} + \binom{2n-2}{r}\binom{n}{1} + \binom{2n-4}{r}\binom{n}{2} + \dots \bigg] $$ now, which I'm not able to..
WolframAlpha doesn't help with that last part. ..or is there a better method to do the whole question?
Working with the innermost sum we have for
$$\sum_{q=0}^n {2n-2q\choose r} {n\choose q}$$
that it is
$$\sum_{q=0}^n [w^r] (1+w)^{2n-2q} [z^{n-q}] (1+z)^n \\ = [w^r] (1+w)^{2n} \sum_{q=0}^n (1+w)^{-2q} [z^{n}] z^q (1+z)^n.$$
Now we may extend $q$ to infinity as there is no contribution to the coefficient extractor $[z^n]$ when $q\gt n,$ getting
$$[w^r] (1+w)^{2n} [z^n] (1+z)^n \sum_{q\ge 0} (1+w)^{-2q} z^q \\ = [w^r] (1+w)^{2n} [z^n] (1+z)^n \frac{1}{1-z/(1+w)^2} \\ = [w^r] (1+w)^{2n+2} [z^n] (1+z)^n \frac{1}{1+2w-z+w^2}.$$
Observe that
$$\sum_{r=0}^{2n} T_r x^r = \sum_{r\ge 0} T_r x^r$$
because we may extend $r$ beyond $2n$ with the coefficient extractor $[w^r]$ returning zero for all $q$ in this case (first line). We obtain
$$\sum_{r\ge 0} x^r (-2)^r 3^n [w^r] (1+w)^{2n+2} [z^n] (1+z)^n \frac{1}{1+2w-z+w^2} \\ = 3^n [z^n] (1+z)^n \sum_{r\ge 0} x^r (-2)^r [w^r] (1+w)^{2n+2} \frac{1}{1+2w-z+w^2} \\ = 3^n [z^n] (1+z)^n (1-2x)^{2n+2} \frac{1}{1-4x-z+4x^2} \\ = 3^n (1-2x)^{2n+2} [z^n] (1+z)^n \frac{1}{1-4x-z+4x^2}.$$
This is
$$\frac{3^n (1-2x)^{2n+2}}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^n \frac{1}{1-4x-z+4x^2} \; dz.$$
Introducing $z/(1+z) = v$ we get $z = v/(1-v)$ and $dz = 1/(1-v)^2 \; dv$ and hence
$$\frac{3^n (1-2x)^{2n+2}}{2\pi i} \int_{|v|=\gamma} \frac{1}{v^n} \frac{1-v}{v} \frac{1}{1-4x-v/(1-v)+4x^2} \; \frac{1}{(1-v)^2} \; dv \\ = \frac{3^n (1-2x)^{2n+2}}{2\pi i} \int_{|v|=\gamma} \frac{1}{v^{n+1}} \frac{1}{(1-4x)(1-v) -v +4x^2 (1-v)} \; dv \\ = \frac{3^n (1-2x)^{2n+2}}{2\pi i} \int_{|v|=\gamma} \frac{1}{v^{n+1}} \frac{1}{(1-4x+4x^2) - v (2-4x+4x^2 ) } \; dv.$$
With $x$ a formal parameter this evaluates to
$$3^n (1-2x)^{2n+2} \frac{1}{1-4x+4x^2} \frac{(2-4x+4x^2)^n}{(1-4x+4x^2)^n} \\ = 3^n (1-2x)^{2n+2} \frac{1}{(1-2x)^2} \frac{(2-4x+4x^2)^n}{(1-2x)^{2n}} \\ = (6-12x+12x^2)^n.$$
This is the claim.