Let 729, 81, 9, 1, dots be a sequence and P_{ | vedclass

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(A) The sequence is a geometric progression with first term $a = 729 = 3^6$ and common ratio $r = \frac{1}{9} = 3^{-2}$.$P_n$ is the product of the fi

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$73$

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(A) The sequence is a geometric progression with first term $a = 729 = 3^6$ and common ratio $r = \frac{1}{9} = 3^{-2}$.$P_n$ is the product of the first $n$ terms: $P_n = a \cdot ar \cdot ar^2 \dots ar^{n-1} = a^n r^{\frac{n(n-1)}{2}}$.$P_n = (3^6)^n \cdot (3^{-2})^{\frac{n(n-1)}{2}} = 3^{6n} \cdot 3^{-n(n-1)} = 3^{6n - n^2 + n} = 3^{7n - n^2}$.Thus, $(P_n)^{\frac{1}{n}} = 3^{7-n}$.We need to evaluate $2 \sum_{n=1}^{40} 3^{7-n} = 2(3^6 + 3^5 + \dots + 3^{7-40}) = 2(3^6 + 3^5 + \dots + 3^{-33})$.This is a geometric series with $40$ terms, first term $A = 3^6$, and common ratio $R = \frac{1}{3}$.Sum $= 2 \cdot 3^6 \left( \frac{1 - (1/3)^{40}}{1 - 1/3} \right) = 2 \cdot 3^6 \cdot \frac{3^{40}-1}{3^{40}} \cdot \frac{3}{2} = \frac{3^7(3^{40}-1)}{3^{40}} = \frac{3^{40}-1}{3^{33}}$.Comparing with $\frac{3^{\alpha}-1}{3^{\beta}}$, we get $\alpha = 40$ and $\beta = 33$.Since $\gcd(40, 33) = 1$, the value of $\alpha + \beta = 40 + 33 = 73$.

Let $729, 81, 9, 1, \dots$ 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

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