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
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 $
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