Question
Let $729,81,9,1,\ldots$. be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence.
If $2\sum_{n = 1}^{40}\mspace{2mu}\left( P_{n} \right)^{\frac{1}{n}} = \frac{3^{\alpha} - 1}{3^{\beta}}$ and $gcd(\alpha,\beta) = 1$, then $\alpha + \beta$ is equal to
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Solution
$P_{n} = 729.81.9\ldots\ldots$. ( n terms)
$$\begin{matrix} & \ = 3^{6} \cdot 3^{4} \cdot 3^{2}\ldots\ldots \cdot \cdot 3^{- 2n + 8} \\ & P_{n} = 3^{6 + 4 + 2 + \ldots\ldots + ( - 2n + 8)} = 3^{n(7 - n)} \\ & P_{n}^{1/n} = 3^{7 - n} \end{matrix}$$
$\Rightarrow \sum_{n = 1}^{40}\mspace{2mu}\left( P_{n} \right)^{\frac{1}{n}} = 3^{6} + 3^{5} + \ldots.. + (40$ terms $)$
$${= 3^{6}\left\lbrack \frac{1 - \left( \frac{1}{3} \right)^{40}}{1 - \frac{1}{3}} \right\rbrack }{= \frac{3^{6}\left\lbrack 3^{40} - 1 \right\rbrack \times 3^{1}}{3^{40} \times 2} }{\sum\left( P_{n} \right)^{\frac{1}{n}} = \frac{\left( 3^{40} - 1 \right)}{2 \times 3^{33}},\ \alpha = 40}$$
$$\beta = 33$$
$$\alpha + \beta = 73$$
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