Given `lim_(n->oo)((.^(3n)C_n)/(.^(2n)C_n))^(1/n) =a/b` where a and b are relatively prime, find the value of `(a + b)`.

Explanation

To solve the problem, we need to evaluate the limit:

\[ \lim_{n \to \infty} \left( \frac{C(3n, n)}{C(2n, n)} \right)^{\frac{1}{n}} \]

where \( C(n, r) \) is the binomial coefficient defined as:

\[ C(n, r) = \frac{n!}{r!(n-r)!} \]

### Step 1: Write the limit in terms of factorials

Using the definition of binomial coefficients, we can express the limit as:

\[ C(3n, n) = \frac{(3n)!}{n!(3n-n)!} = \frac{(3n)!}{n!(2n)!} \]

\[ C(2n, n) = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!} \]

Thus, we can rewrite the limit as:

\[ \lim_{n \to \infty} \left( \frac{(3n)!}{n!(2n)!} \cdot \frac{n!n!}{(2n)!} \right)^{\frac{1}{n}} = \lim_{n \to \infty} \left( \frac{(3n)!}{(2n)! \cdot n!} \right)^{\frac{1}{n}} \]

### Step 2: Simplify the expression

This simplifies to:

\[ \lim_{n \to \infty} \left( \frac{(3n)!}{(2n)! \cdot n!} \right)^{\frac{1}{n}} \]

### Step 3: Apply Stirling's approximation

Using Stirling's approximation, \( n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \), we can approximate the factorials:

\[ (3n)! \sim \sqrt{6 \pi n} \left( \frac{3n}{e} \right)^{3n} \]
\[ (2n)! \sim \sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n} \]
\[ n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \]

### Step 4: Substitute the approximations

Substituting these approximations into our limit gives:

\[ \lim_{n \to \infty} \left( \frac{\sqrt{6 \pi n} \left( \frac{3n}{e} \right)^{3n}}{\sqrt{4 \pi n} \left( \frac{2n}{e} \right)^{2n} \cdot \sqrt{2 \pi n} \left( \frac{n}{e} \right)^{n}} \right)^{\frac{1}{n}} \]

### Step 5: Simplify the limit

This simplifies to:

\[ \lim_{n \to \infty} \left( \frac{\sqrt{6}}{\sqrt{8 \pi n}} \cdot \frac{(3n)^{3n}}{(2n)^{2n} \cdot n^n} \cdot e^{n}} \right)^{\frac{1}{n}} \]

The dominant term in the limit is:

\[ \frac{(3n)^{3n}}{(2n)^{2n} \cdot n^n} = \frac{3^{3n} n^{3n}}{2^{2n} n^{2n} n^n} = \frac{3^{3n}}{2^{2n}} = \left( \frac{27}{4} \right)^{n} \]

### Step 6: Evaluate the limit

Thus, we have:

\[ \lim_{n \to \infty} \left( \frac{27}{4} \right) = \frac{27}{4} \]

### Step 7: Identify \( a \) and \( b \)

Here, \( a = 27 \) and \( b = 4 \). Since 27 and 4 are relatively prime, we find:

\[ a + b = 27 + 4 = 31 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{31} \]