Limit of quotient of summations involving special binomial coefficients Find the limit, when $n$ tends to infinity, of
$$
  \frac{\displaystyle\sum_{k=0}^n\binom{2n}{2k}3^k}
       {\displaystyle\sum_{k=0}^{n-1}\binom{2n}{2k+1}3^k}
$$
Please Help Me to solve the problem given and i have no idea to do since  i have done such this  one and need help also many hint of course.
 A: Note that
$$
\sum_{k=0}^n\binom{2n}{2k}3^k-\sqrt3\sum_{k=0}^{n-1}\binom{2n}{2k+1}3^k=(\sqrt3-1)^{2n}
$$
and
$$
\sum_{k=0}^n\binom{2n}{2k}3^k+\sqrt3\sum_{k=0}^{n-1}\binom{2n}{2k+1}3^k=(\sqrt3+1)^{2n}
$$
Therefore,
$$
\begin{align}
\frac{\displaystyle\sum_{k=0}^n\binom{2n}{2k}3^k}{\displaystyle\sum_{k=0}^{n-1}\binom{2n}{2k+1}3^k}
&=\sqrt3\,\frac{(\sqrt3+1)^{2n}+(\sqrt3-1)^{2n}}{(\sqrt3+1)^{2n}-(\sqrt3-1)^{2n}}\\
&=\sqrt3\,\frac{1+\left(\frac{\sqrt3-1}{\sqrt3+1}\right)^{2n}}{1-\left(\frac{\sqrt3-1}{\sqrt3+1}\right)^{2n}}
\end{align}
$$
Taking the limit is pretty simple.
A: HINT: 
$\displaystyle\sum_{k=0}^n\binom{2n}{2k}3^k=\sum_{k=0}^n\binom{2n}{2k}(\sqrt3)^{2k}$
and $\displaystyle\sum_{k=0}^{n-1}\binom{2n}{2k+1}3^k=\dfrac{\sum_{k=0}^{n-1}\binom{2n}{2k+1}(\sqrt3)^{2k+1}}{\sqrt3}$
Use $(1+x)^{2n}+(1-x)^{2n}=\cdots$ and $(1+x)^{2n}-(1-x)^{2n}=\cdots$
A: Hint: You can use that for any $m$, and $i\in\{0,1\}$ one has
$$
  \sum_{k\equiv i\pmod2}\binom mkX^k
 =\frac12\sum_k(1^k+(-1)^{k+i})\binom mkX^k
 =\frac12((1+X)^m+(-1)^i(1-X)^m),
$$
and apply this with $m=2n$ and $X=\sqrt3$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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${\ds{\lim_{n\ \to\ \infty}}\ {\ds{\sum_{k\ =\ 0}^{n}\ {2n \choose 2k}3^{k}}\over
  \ds{\sum_{k\ =\ 0}^{n - 1}\ {2n \choose 2k + 1}3^{k}}}: \ {\large ?}}$.

The expression $\ds{{2n \choose 2k + s}3^{k}}$, with
$\ds{s = 0, 1}$, is 'highly concentrated' around
$\ds{\tilde{k}_{s} = {6n - 3s - \root{3} s \over 2\pars{3 + \root{3}}}}$ when
$\ds{n \gg 1}$such that
$$
\sum_{k\ =\ 0}^{n - s}{2n \choose 2k + s}3^{k}
\sim {2n \choose 2\tilde{k}_{s} + s }3^{\tilde{k}_{s}}
\int_{0}^{n - s}
\exp\pars{-\,{\bracks{k - \tilde{k}_{s}}^{2} \over 2\sigma^{2}}}\,\dd k\,,
\qquad\sigma\equiv{\root{3} \over 2\root{3 + 2\root{3}}}\,n^{1/2}
$$

Then,
\begin{align}
{\ds{\sum_{k\ =\ 0}^{n}\ {2n \choose 2k}3^{k}}\over
 \ds{\sum_{k\ =\ 0}^{n - 1}\ {2n \choose 2k + 1}3^{k}}}
\sim{\ds{2n \choose 2\tilde{k}_{0}}3^{\tilde{k}_{0}} \over \ds{2n \choose 2\tilde{k}_{1} + 1}3^{\tilde{k}_{1}}}
=3^{\tilde{k}_{0} - \tilde{k}_{1}}=\root{3}
\qquad n \gg 1
\end{align}
and
\begin{align}\color{#66f}{\large\lim_{n\ \to\ \infty}\
{\ds{\sum_{k\ =\ 0}^{n}\ {2n \choose 2k}3^{k}}\over
 \ds{\sum_{k\ =\ 0}^{n - 1}\ {2n \choose 2k + 1}3^{k}}}}
=\color{#66f}{\large\root{3}}
\end{align}
