Let $$f^1(x) = \frac{3x+2}{2x+3}$$, $$x \in R - \{-\frac{3}{2}\}$$. For $$n \geq 2$$, define $$f^n x = f^1 \circ f^{n-1}(x)$$. If $$f^5 x = \frac{ax+b}{bx+a}$$, $$\gcd(a,b) = 1$$, then $$a + b$$ is equal to ______.

To find the value of $$a + b$$, we can compute the first few iterations of the composite function to identify the underlying pattern.

The first iteration is given as:

$$f^1(x) = \frac{3x+2}{2x+3}$$

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Step 1: Compute the second iteration $$f^2(x)$$

$$f^2(x) = f^1(f^1(x)) = \frac{3\left(\frac{3x+2}{2x+3}\right) + 2}{2\left(\frac{3x+2}{2x+3}\right) + 3}$$

Multiplying the numerator and denominator by $$(2x+3)$$ to clear the fractions:

$$f^2(x) = \frac{3(3x+2) + 2(2x+3)}{2(3x+2) + 3(2x+3)} = \frac{9x + 6 + 4x + 6}{6x + 4 + 6x + 9} = \frac{13x + 12}{12x + 13}$$

Notice that the sum of the coefficients in the numerator of $$f^2(x)$$ is $$13 + 12 = 25 = 5^2$$.

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Step 2: Compute the third iteration $$f^3(x)$$

$$f^3(x) = f^1(f^2(x)) = \frac{3\left(\frac{13x+12}{12x+13}\right) + 2}{2\left(\frac{13x+12}{12x+13}\right) + 3}$$

Multiplying the numerator and denominator by $$(12x+13)$$:

$$f^3(x) = \frac{3(13x+12) + 2(12x+13)}{2(13x+12) + 3(12x+13)} = \frac{39x + 36 + 24x + 26}{26x + 24 + 36x + 39} = \frac{63x + 62}{62x + 63}$$

Notice that the sum of the coefficients in the numerator of $$f^3(x)$$ is $$63 + 62 = 125 = 5^3$$.

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Step 3: Generalize the pattern for $$f^n(x)$$

From the observed results, any iteration $$f^n(x)$$ can be represented in the symmetric form:

$$f^n(x) = \frac{a_n x + b_n}{b_n x + a_n}$$

Furthermore, the sum of the coefficients in the numerator satisfies the following exponential relation:

$$a_n + b_n = 5^n$$

Additionally, by tracking the difference of the coefficients ($$a_1 - b_1 = 3 - 2 = 1$$, $$a_2 - b_2 = 13 - 12 = 1$$, etc.), we have $$a_n - b_n = 1$$ for all $$n$$. Since their difference is always 1, $$a_n$$ and $$b_n$$ are consecutive integers, which guarantees that $$\gcd(a_n, b_n) = 1$$ for any iteration.

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Step 4: Calculate $$a + b$$ for $$f^5(x)$$

For $$n = 5$$, the coefficients are $$a$$ and $$b$$ such that $$f^5(x) = \frac{ax+b}{bx+a}$$ with $$\gcd(a,b) = 1$$. Following our general relation:

$$a + b = 5^5$$

Therefore, the value of $$a + b$$ is equal to 3125.