Limit of a ratio For a positive integer $n$, let $a_n, b_n, c_n, d_n $ be positive integers such that
$$\left(1+\sqrt 2+\sqrt 3\right)^n=a_n+b_n \sqrt 2+c_n\sqrt 3+d_n\sqrt 6$$
Then for large $ n$, find the limit of the expression
$$\frac{a_n^2+b_n^2+c_n^2+d_n^2}{(a_n+b_n+c_n+d_n)^2}$$ I tried some binomial theorem for the expansion, but no progress.
 A: Please check my simplification work to get $I$:
$$A=(1+\sqrt{2}+\sqrt{3})^n=a_{n}+b_{n}\sqrt{2}+c_{n}\sqrt{3}+d_{n}\sqrt{6}$$
$$B=(1-\sqrt{2}+\sqrt{3})^n=a_{n}-b_{n}\sqrt{2}+c_{n}\sqrt{3}-d_{n}\sqrt{6}$$
$$C=(1+\sqrt{2}-\sqrt{3})^n=a_{n}+b_{n}\sqrt{2}-c_{n}\sqrt{3}-d_{n}\sqrt{6}$$
$$D=(1-\sqrt{2}-\sqrt{3})^n=a_{n}-b_{n}\sqrt{2}-c_{n}\sqrt{3}+d_{n}\sqrt{6}$$
then we have
$$4a_{n}=A^n+B^n+C^n+D^n,$$
$$4\sqrt{2}b_{n}=A^n-B^n+C^n-D^n,$$
$$4\sqrt{3}c_{n}=A^n+B^n-C^n-D^n,$$
$$ \sqrt{6}d_{n}=A^n-B^n-C^n+D^n,$$
then it is easy to find
$$x=\lim_{n\to\infty}\dfrac{b_{n}}{a_{n}}=\lim_{n\to\infty}\dfrac{\dfrac{1}{4\sqrt{2}}(A^n-B^n+C^n-D^n)}{\dfrac{1}{4}(A^n+B^n+C^n+D^n}=\dfrac{\sqrt{2}}{2}$$
and similarly
$$y=\lim_{n\to\infty}\dfrac{c_{n}}{a_{n}}=\dfrac{1}{\sqrt{3}},z=\lim_{n\to\infty}\dfrac{d_{n}}{a_{n}}=\dfrac{1}{\sqrt{6}}$$
So
$$I=\lim_{n\to\infty}\dfrac{a^2_{n}+b^2_{n}+c^2_{n}+d^2_{n}}{(a_{n}+b_{n}+c_{n}+d_{n})^2}
=\dfrac{1+x^2+y^2+z^2}{(1+x+y+z)^2}=\dfrac{72}{(6+3\sqrt{2}+2\sqrt{3}+\sqrt{6})^2}$$
and note
$$6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}=3\sqrt{2}(1+\sqrt{2})+\sqrt{6}(1+\sqrt{2})=(1+\sqrt{2})(3\sqrt{2}+\sqrt{6})=\sqrt{6}(1+\sqrt{2})^2$$
so
$$I=\dfrac{12}{(1+\sqrt{2})^4}=12(\sqrt{2}-1)^4$$
