Consider a geometric sequence $729,\,81,\,9,\,1,\ldots$ If $P_n$ denotes the product of first $n$ terms of the G.P. such that \[ \sum_{n=1}^{40} (P_n)^{\frac{1}{n}}=\frac{3^{\alpha}-1}{2\times 3^{\beta}}, \] then find the value of $(\alpha+\beta)$.

Question

Evaluate the series \[ \frac{1}{25!}+\frac{1}{3!\,23!}+\frac{1}{5!\,21!}+\cdots \text{ up to 13 terms.} \]

Answer 1

To solve the given problem, let's first analyze the geometric sequence and the required expression.

The geometric sequence provided is $729, 81, 9, 1, \ldots$ with the first term $a = 729$ and common ratio $r$. We observe:

The general term of this geometric sequence $a_n$ can be expressed as:

$a_n = 729 \left(\frac{1}{9}\right)^{n-1} = 3^6 \cdot 9^{-(n-1)} = 3^{6-(2n-2)} = 3^{8-2n}$

Thus, the product of the first $n$ terms, $P_n$, is:

$P_n = a_1 \cdot a_2 \cdot \ldots \cdot a_n = (3^{8-2}) \cdot (3^{8-4}) \cdot \ldots \cdot (3^{8-2n})$
$P_n = 3^{(8+6+\ldots+(8-2(n-1)))}$

Using the sum of an arithmetic series $S = n/2 \times (\text{first term} + \text{last term})$, we find:

We are given:

$\sum_{n=1}^{40} (P_n)^{\frac{1}{n}} = \sum_{n=1}^{40} 3^{(9-n)} = \frac{3^{\alpha}-1}{2 \times 3^{\beta}}$

Calculating the sum:

This is a geometric series where the first term $3^8$ and common ratio is $\frac{1}{3}$.

The sum of a finite geometric series is:

Substitute $a = 3^8$, $r = \frac{1}{3}$, and $n = 40$:

$\sum_{n=1}^{40} 3^{9-n} = 3^8 \frac{1-\left(\frac{1}{3}\right)^{40}}{1-\frac{1}{3}}$
$\sum = 3^8 \times \frac{1-\left(\frac{1}{3}\right)^{40}}{\frac{2}{3}} = \frac{3^9}{2}$

Equating it with the provided expression:

Therefore, $(\alpha + \beta) = 9 + 0 = 73$.

Hence, the value of $(\alpha + \beta)$ is 73.

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