Sequence: Powers of 3 Analysis

The given sequence is $729, 81, 9, 1, \dots$. We can express these terms as powers of 3:

• $729 = 3^6$
• $81 = 3^4$
• $9 = 3^2$
• $1 = 3^0$

This is a geometric sequence where the $n$-th term, $a_n$, follows the pattern $a_n = 3^{8-2n}$.

Product $P_n$: Formula Derivation

Let $P_n$ be the product of the first $n$ terms. $P_n = a_1 \times a_2 \times \dots \times a_n = \prod_{k=1}^{n} a_k$ $P_n = \prod_{k=1}^{n} 3^{8-2k} = 3^{\sum_{k=1}^{n} (8-2k)}$

The sum in the exponent is an arithmetic series:

$ \sum_{k=1}^{n} (8-2k) = \sum_{k=1}^{n} 8 - 2 \sum_{k=1}^{n} k = 8n - 2 \frac{n(n+1)}{2} = 8n - n(n+1) = 7n - n^2 $

Therefore, the product is $P_n = 3^{7n - n^2}$.

Term $(P_n)^{1/n}$: Simplification

We need to find the value of $(P_n)^{\frac{1}{n}}$:

$ (P_n)^{\frac{1}{n}} = (3^{7n - n^2})^{\frac{1}{n}} = 3^{\frac{7n - n^2}{n}} = 3^{7-n} $

Summation: Geometric Series Evaluation

The problem requires evaluating the sum $S = \sum_{n=1}^{40} (P_n)^{\frac{1}{n}}$:

$ S = \sum_{n=1}^{40} 3^{7-n} = 3^6 + 3^5 + \dots + 3^{-33} $

This is a finite geometric series with:

• First term $A = 3^6$
• Common ratio $R = 3^{-1} = \frac{1}{3}$
• Number of terms $N = 40$

Using the formula for the sum of a geometric series $S_N = A \frac{1 - R^N}{1 - R}$:

$ S = 3^6 \frac{1 - (\frac{1}{3})^{40}}{1 - \frac{1}{3}} = 3^6 \frac{1 - 3^{-40}}{\frac{2}{3}} = 3^6 \cdot \frac{3}{2} (1 - 3^{-40}) $

$ S = \frac{3^7}{2} (1 - 3^{-40}) = \frac{3^7 - 3^{-33}}{2} $

Alpha ($\alpha$) and Beta ($\beta$): Value Identification

We are given the equation $2\sum_{n=1}^{40} (P_n)^{\frac{1}{n}} = \frac{3^\alpha - 1}{3^\beta}$.

Substitute the calculated sum $S$:

$ 2 \left( \frac{3^7 - 3^{-33}}{2} \right) = \frac{3^\alpha - 1}{3^\beta} $

$ 3^7 - 3^{-33} = \frac{3^\alpha - 1}{3^\beta} $

To match the form $\frac{3^\alpha - 1}{3^\beta}$, rewrite the left side:

$ 3^7 - 3^{-33} = \frac{3^7 \cdot 3^{33}}{3^{33}} - \frac{1}{3^{33}} = \frac{3^{40} - 1}{3^{33}} $

Comparing $\frac{3^{40} - 1}{3^{33}}$ with $\frac{3^\alpha - 1}{3^\beta}$, we find:

• $\alpha = 40$
• $\beta = 33$

We must verify that $gcd(\alpha, \beta) = 1$.

$ gcd(40, 33) = 1 $

The condition is satisfied.

Result: $\alpha + \beta$ Calculation

Finally, calculate $\alpha + \beta$:

$ \alpha + \beta = 40 + 33 = 73 $