Let 729, 81 , 9, 1, ... be a sequence and $$P_{n}$$ denote the product of the first n terms of this sequence.
If $$2\sum_{n=1}^{40}(P_{n})^{\frac{1}{n}}=\frac{3^{\alpha}-1}{3^{\beta}}$$ and $$gcd(\alpha\beta)=1$$ then $$\alpha+\beta$$ is equal to

The sequence is 729, 81, 9, 1, ... We need to find $$\alpha + \beta$$ where $$2\sum_{n=1}^{40}(P_n)^{1/n} = \frac{3^\alpha - 1}{3^\beta}$$ and $$\gcd(\alpha, \beta) = 1$$.

First, identify the sequence by observing that $$729 = 3^6, \; 81 = 3^4, \; 9 = 3^2, \; 1 = 3^0, \ldots$$ It is a geometric progression with first term $$a = 3^6$$ and common ratio $$r = 3^{-2} = 1/9$$. Therefore, the general term is $$a_n = 3^{6-2(n-1)} = 3^{8-2n}$$.

Next, the product of the first n terms is $$P_n = \prod_{k=1}^{n} 3^{8-2k} = 3^{\sum_{k=1}^{n}(8-2k)}$$. Noting that $$\sum_{k=1}^{n}(8-2k) = 8n - 2 \cdot \frac{n(n+1)}{2} = 8n - n(n+1) = n(7-n)$$, we obtain $$P_n = 3^{n(7-n)}$$.

Taking the nth root gives $$(P_n)^{1/n} = 3^{7-n}$$.

Evaluating the sum, we have:

$$2\sum_{n=1}^{40} 3^{7-n} = 2 \cdot 3^7 \sum_{n=1}^{40} 3^{-n} = 2 \cdot 3^7 \cdot \frac{3^{-1}(1 - 3^{-40})}{1 - 3^{-1}} = 2 \cdot 3^7 \cdot \frac{\frac{1}{3}(1 - 3^{-40})}{\frac{2}{3}} = 2 \cdot 3^7 \cdot \frac{1}{2}(1 - 3^{-40}) = 3^7(1 - 3^{-40}) = 3^7 - 3^{-33} = \frac{3^{40} - 1}{3^{33}}$$.

Comparing this with the form $$\frac{3^\alpha - 1}{3^\beta}$$ shows that $$\alpha = 40$$ and $$\beta = 33$$. Since $$\gcd(40, 33) = 1$$, it follows that $$\alpha + \beta = 40 + 33 = 73$$.

The correct answer is Option 1: 73.