Find the limit, when $n$ tends to the infinity, of $$\frac{\sum_{k=0}^{n} {{2n} \choose {2k}} 3^k} {\sum_{k=0}^{n-1} {{2n} \choose {2k+1}} 3^k}$$
To find the limit of the given expression as \( n \) tends to infinity, we start by analyzing the sums in the numerator and the denominator using the Binomial Theorem.

1. **Numerator Analysis:**
   The numerator is given by:
   \[
   \sum_{k=0}^{n} \binom{2n}{2k} 3^k
   \]
   Using the Binomial Theorem, we can express this sum as:
   \[
   \sum_{k=0}^{n} \binom{2n}{2k} 3^k = \frac{(1+\sqrt{3})^{2n} + (1-\sqrt{3})^{2n}}{2}
   \]
   This is because the even terms in the binomial expansion of \((1+\sqrt{3})^{2n}\) and \((1-\sqrt{3})^{2n}\) contribute to the sum.

2. **Denominator Analysis:**
   The denominator is given by:
   \[
   \sum_{k=0}^{n-1} \binom{2n}{2k+1} 3^k
   \]
   Similarly, using the Binomial Theorem, we can express this sum as:
   \[
   \sum_{k=0}^{n-1} \binom{2n}{2k+1} 3^k = \frac{1}{\sqrt{3}} \cdot \frac{(1+\sqrt{3})^{2n} - (1-\sqrt{3})^{2n}}{2}
   \]
   This is because the odd terms in the binomial expansion of \((1+\sqrt{3})^{2n}\) and \((1-\sqrt{3})^{2n}\) contribute to the sum.

3. **Forming the Limit Expression:**
   Now, we form the limit expression:
   \[
   \lim_{n \to \infty} \frac{\sum_{k=0}^{n} \binom{2n}{2k} 3^k}{\sum_{k=0}^{n-1} \binom{2n}{2k+1} 3^k}
   \]
   Substituting the expressions derived from the Binomial Theorem, we get:
   \[
   \lim_{n \to \infty} \frac{\frac{(1+\sqrt{3})^{2n} + (1-\sqrt{3})^{2n}}{2}}{\frac{1}{\sqrt{3}} \cdot \frac{(1+\sqrt{3})^{2n} - (1-\sqrt{3})^{2n}}{2}}
   \]

4. **Simplifying the Expression:**
   Simplify the fraction:
   \[
   \lim_{n \to \infty} \frac{(1+\sqrt{3})^{2n} + (1-\sqrt{3})^{2n}}{(1/\sqrt{3}) \cdot ((1+\sqrt{3})^{2n} - (1-\sqrt{3})^{2n})}
   \]
   \[
   = \lim_{n \to \infty} \frac{\sqrt{3} \cdot ((1+\sqrt{3})^{2n} + (1-\sqrt{3})^{2n})}{(1+\sqrt{3})^{2n} - (1-\sqrt{3})^{2n}}
   \]

5. **Analyzing the Dominant Term:**
   As \( n \) tends to infinity, \((1+\sqrt{3})^{2n}\) grows exponentially while \((1-\sqrt{3})^{2n}\) tends to zero because \(|1-\sqrt{3}| < 1\). Therefore, the dominant term in both the numerator and the denominator is \((1+\sqrt{3})^{2n}\).

6. **Final Limit Calculation:**
   \[
   \lim_{n \to \infty} \frac{\sqrt{3} \cdot (1+\sqrt{3})^{2n}}{(1+\sqrt{3})^{2n}} = \sqrt{3}
   \]

Thus, the limit is \(\sqrt{3}\).

The final answer is \(\boxed{\sqrt{3}}\).