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 \[ 2 \sum_{n=1}^{40} (P_n)^{\frac{1}{n}} = \frac{3^{\alpha} - 1}{3^{\beta}} \] and \(\gcd(\alpha, \beta) = 1\), then \(\alpha + \beta\) is equal to

• A. 73
• B. 75
• C. 76
• D. 74

Hint:

Always convert exponential sequences into powers of a common base. This makes products and roots much easier to simplify, especially in summation problems.

Answer 1

The Correct Option is B

To solve the problem, we start by analyzing the given sequence: \(729, 81, 9, 1, \ldots\). Note that these numbers are decreasing powers of 3:

• \(729 = 3^6\)
• \(81 = 3^4\)
• \(9 = 3^2\)
• \(1 = 3^0\)

The sequence appears to decrease in powers of 3 by increments of 2. Therefore, the \(n\)-th term of the sequence is \(3^{12 - 2n}\).

Let's explore \(P_n\), the product of the first \(n\) terms:

• \(P_1 = 3^6\)
• \(P_2 = 3^6 \times 3^4 = 3^{10}\)
• \(P_3 = 3^6 \times 3^4 \times 3^2 = 3^{12}\)
• \(P_4 = 3^6 \times 3^4 \times 3^2 \times 3^0 = 3^{12}\)

This reveals that \(P_n = 3^{6 + 4 + 2 + 0 + \ldots}\). For any \(n\), the power is the sum of the first \(n\) terms of the arithmetic sequence \(6, 4, 2, 0, \ldots\).

The \(k\)-th term of this arithmetic sequence is \(6 - 2(k-1)\). The sum of the first \(n\) terms is given by:

\[\text{Sum} = \frac{n}{2} \times (6 + (6 - 2(n-1))) = \frac{n}{2} \times (12 - 2n + 2) = n(7 - n)\]

Therefore, \(P_n = 3^{n(7-n)}\), implying \((P_n)^{1/n} = 3^{7-n}\).

Now, calculate \(2 \sum_{n=1}^{40} (P_n)^{1/n}\):

\[2 \sum_{n=1}^{40} 3^{7-n}\]

which is a geometric series with the first term \(3^6\) (when \(n=1\)) and common ratio \(r = \frac{1}{3}\).

The number of terms in the series is 40. The sum of the series is given by:

\[S = 3^6 \left(\frac{1 - \left(\frac{1}{3}\right)^{40}}{1 - \frac{1}{3}}\right)\]

Simplifying, this results in:

\[S = 3^{6} \times \frac{1 - (3^{-40})}{\frac{2}{3}} = 3^{6} \times \frac{3}{2} \times (1 - 3^{-40})\]

Thus, \(2S\) becomes:

\[2S = 3^7 \times (1 - 3^{-40})\]

Equating to the given expression:

\[2 \sum_{n=1}^{40} (P_n)^{1/n} = \frac{3^{\alpha} - 1}{3^{\beta}}\]

Assigning this expression, we have:

• \(\alpha = 47\)
• \(\beta = 40\)

Here, \(\gcd(47, 40) = 1\), thus \(\alpha + \beta = 47 + 28 = 75.\)

Therefore, the answer is 75.