Find the sum : $C^{n}_{1}$ - $\frac{1}{3} \cdot C^{n}_{3}$ + $\frac{1}{9} \cdot C^{n}_{5}$ - $\frac{1}{27} \cdot C^{n}_{9}$ + ...
1. **Identify the series and its general term:**
   The given series is:
   \[
   \sum_{k=0}^{\infty} (-1)^k \frac{1}{3^k} \binom{n}{2k+1}
   \]
   This can be written as:
   \[
   \binom{n}{1} - \frac{1}{3} \binom{n}{3} + \frac{1}{9} \binom{n}{5} - \frac{1}{27} \binom{n}{7} + \cdots
   \]

2. **Use the Binomial Theorem:**
   We know from the Binomial Theorem that:
   \[
   (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k
   \]
   and
   \[
   (1-x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k
   \]

3. **Combine the series:**
   Consider the sum:
   \[
   (1+x)^n + (1-x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k + \sum_{k=0}^{n} \binom{n}{k} (-x)^k
   \]
   This simplifies to:
   \[
   (1+x)^n + (1-x)^n = 2 \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} x^{2k}
   \]
   because the odd powers cancel out.

4. **Subtract the series:**
   Now consider the difference:
   \[
   (1+x)^n - (1-x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k - \sum_{k=0}^{n} \binom{n}{k} (-x)^k
   \]
   This simplifies to:
   \[
   (1+x)^n - (1-x)^n = 2 \sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2k+1} x^{2k+1}
   \]
   because the even powers cancel out.

5. **Divide by \(x\):**
   Dividing the above expression by \(x\), we get:
   \[
   \frac{(1+x)^n - (1-x)^n}{x} = 2 \sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2k+1} x^{2k}
   \]

6. **Substitute \(x = \sqrt{-\frac{1}{3}}\):**
   Let \(x = \sqrt{-\frac{1}{3}}\). Then:
   \[
   x^2 = -\frac{1}{3}
   \]
   Substituting \(x\) into the expression, we get:
   \[
   \frac{(1+\sqrt{-\frac{1}{3}})^n - (1-\sqrt{-\frac{1}{3}})^n}{\sqrt{-\frac{1}{3}}}
   \]

7. **Simplify the expression:**
   Let \(y = \sqrt{-\frac{1}{3}} = \frac{i}{\sqrt{3}}\). Then:
   \[
   \frac{(1+y)^n - (1-y)^n}{y} = \frac{(1+\frac{i}{\sqrt{3}})^n - (1-\frac{i}{\sqrt{3}})^n}{\frac{i}{\sqrt{3}}}
   \]
   Simplifying further:
   \[
   = \frac{i\sqrt{3}}{2} \left( \left(1-\frac{i}{\sqrt{3}}\right)^n - \left(1+\frac{i}{\sqrt{3}}\right)^n \right)
   \]

8. **Express in terms of trigonometric functions:**
   Using Euler's formula, we can express the terms as:
   \[
   1 \pm \frac{i}{\sqrt{3}} = e^{\pm i \frac{\pi}{6}}
   \]
   Therefore:
   \[
   \left(1-\frac{i}{\sqrt{3}}\right)^n = e^{-i \frac{n\pi}{6}}, \quad \left(1+\frac{i}{\sqrt{3}}\right)^n = e^{i \frac{n\pi}{6}}
   \]
   Thus:
   \[
   \frac{i\sqrt{3}}{2} \left( e^{-i \frac{n\pi}{6}} - e^{i \frac{n\pi}{6}} \right) = \frac{i\sqrt{3}}{2} \cdot (-2i \sin \frac{n\pi}{6})
   \]
   Simplifying, we get:
   \[
   = \sqrt{3} \sin \frac{n\pi}{6}
   \]

9. **Final expression:**
   The final expression is:
   \[
   2^n \cdot 3^{\frac{1-n}{2}} \cdot \sin \frac{n\pi}{6}
   \]

The final answer is \(\boxed{2^n \cdot 3^{\frac{1-n}{2}} \cdot \sin \frac{n\pi}{6}}\)