Binomial Theorem proof using three terms

Question

If $$(6-12x+12x^2)^n =\sum\limits_{r=0}^{2n}a_{r}x^r$$ then prove that $$a_{r} = (-1)^r3^n2^r\left[{2n \choose r} + {n \choose 1}{(2n - 2) \choose r} + {n \choose 2}{(2n-4) \choose r} + \cdots \right]$$

I tried to expand the binomial by grouping the first two terms as one term and the third term as second term. $$ \begin{split} (6-12x+12x^2)^n & = \big((6-12x)+12x^2\big)^n \\ &= (6-12x)^n + {n \choose 1}(6-12x)^{(n-1)}12x^2 \\ &\quad+ {n \choose 2}(6-12x)^{(n-2)}\big(12x^2\big)^2 + {n \choose 3}(6-12x)^{(n-3)}\big(12x^2\big)^3 + \ldots + {n \choose n}\big(12x^2\big)^n. \end{split} $$ Then I extracted the power of $x^r$: $${n \choose r}6^{n-r}(-12)^r + {n \choose 1}{n-1 \choose r-2}6^{n-r+1}(-12)^{r-2} + {n \choose 2}{n-2 \choose r-4}6^{n-r+2}(-12)^{r-4} + \ldots $$ After this step I was clueless how to proceed to achieve the right hand side of equality and conclude the proof.

Answer 1

Even if we were not given the value of the $a_r$s, we could discover it (and thereby prove it), by expanding $(6-12x+12x^2)^n$ with two consecutive uses of the binomial theorem, and a change of the order of summation: $$\begin{align}(6-12x+12x^2)^n&=3^n\left(1+(2x-1)^2\right)^n\\&=3^n\sum_{k=0}^n\binom nk(2x-1)^{2(n-k)}\\&=3^n\sum_{k=0}^n\binom nk\sum_{r=0}^{2(n-k)}\binom{2(n-k)}r (2x)^r(-1)^{2(n-k)-r}\\&=\sum_{r=0}^{2n}a_rx^r \end{align}$$ with $$a_r=3^n(-2)^r\sum_{k=0}^n\binom nk\binom{2(n-k)}r.$$

Note that all along these calculations, the lower and upper bounds in the sums were actually superfluous. We could as well write $\sum_{k\in\Bbb N}$ and $\sum_{r\in\Bbb N}$, with the convention that $\binom pq=0$ if $q>p$.

More generally, you can use this method to prove that $$((ax+b)^2+c)^n=\sum_{r=0}^{2n}a_rx^r$$ with $$a_r=b^{2n}\left(\frac ab\right)^r\sum_{k=0}^n\binom nk\binom{2(n-k)}r\left(\frac c{b^2}\right)^k.$$

Answer 2

Induction:

$(6-12x+12x^2)^n =\sum\limits_{r=0}^{2n}a_{r}x^r$

So $(6-12x + 12x^2)^{n+1} = (6 - 12x +12x^2)\sum\limits_{r=0}^{2n}a_{r}x^r=$

$6\sum\limits_{r=0}^{2n}a_{r}x^r + 12\sum\limits_{r=1}^{2n+1}a_{r-1}x^r+ 12\sum\limits_{r=2}^{2n+2}a_{r-2}x^r$

So if $(6-12x + 12x^2)^{n+1}=\sum\limits_{r=0}^{2(n+1)}b_rx^r$

we have $b_0=6a_0; b_1 = 6a_1 + 12a_0$ and for $2\le r \le 2n$ we have $b_r=6a_r +12a_{r-1} + 12_{r-2}$ and $b_{2n+1} = 12a_{2n} + 12a+{2n-1}$ and $b_{2n+2}= 12a_{2n}$.

Plug in $a_{r} = (-1)^r.3^n.2^r.({2n \choose r} + {n \choose 1}.{(2n - 2) \choose r} + {n \choose 2}. {(2n-4) \choose r} + ... )$

And if we we are correct, this should yield:

$b_r = (-1)^r.3^{n+1}.2^r.({2n+2 \choose r} + {n+1 \choose 1}.{2n \choose r} + {n+1 \choose 2}. {(2n-2) \choose r} + ... )$