(A) 73

(B) 74

(C) 75

(D) 76

Correct Answer: 73

Step-by-Step Solution

Given sequence:

$$ 729,\,81,\,9,\,1,\ldots = 3^6,\,3^4,\,3^2,\,3^0,\ldots $$

Hence, the $n^{\text{th}}$ term is:

$$ a_n = 3^{\,6-2(n-1)} = 3^{\,8-2n} $$

Product of first $n$ terms:

$$ P_n = \prod_{k=1}^{n} 3^{\,8-2k} = 3^{\sum_{k=1}^{n}(8-2k)} $$

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

$$ \Rightarrow P_n = 3^{\,7n-n^2} $$

Now,

$$ (P_n)^{1/n} = 3^{\,7-n} $$

So the summation becomes:

$$ 2\sum_{n=1}^{40}3^{\,7-n} $$

This is a geometric series:

$$ \sum_{n=1}^{40}3^{7-n} =3^6\left(\frac{1-(1/3)^{40}}{1-1/3}\right) =\frac{3^7}{2}\left(1-\frac{1}{3^{40}}\right) $$

Multiplying by 2:

$$ 2\sum_{n=1}^{40}(P_n)^{1/n} =3^7\left(1-\frac{1}{3^{40}}\right) =\frac{3^{47}-3^7}{3^{40}} $$

Comparing with

$$ \frac{3^{\alpha}-1}{3^{\beta}} $$

we get:

$$ \alpha=47,\quad \beta=40 $$

$$ \alpha+\beta=47+40=\boxed{73} $$

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