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
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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

Step 1: Write the general term of the sequence. 
The given sequence is \[ 729, 81, 9, 1, \ldots \] which can be written as powers of 3: \[ 729 = 3^6,\quad 81 = 3^4,\quad 9 = 3^2,\quad 1 = 3^0 \] Hence, the \(n\)th term is: \[ a_n = 3^{2(4-n)} \] Step 2: Find the product \(P_n\) of the first \(n\) terms. 
\[ P_n = \prod_{k=1}^{n} 3^{2(4-k)} = 3^{2 \sum_{k=1}^{n} (4-k)} \] \[ \sum_{k=1}^{n} (4-k) = 4n - \frac{n(n+1)}{2} \] So, \[ P_n = 3^{\,2\left(4n - \frac{n(n+1)}{2}\right)} = 3^{\,8n - n(n+1)} \] Step 3: Evaluate \((P_n)^{1/n}\). 
\[ (P_n)^{\frac{1}{n}} = 3^{\frac{8n - n(n+1)}{n}} = 3^{7-n} \] Step 4: Evaluate the given summation. 
\[ 2 \sum_{n=1}^{40} 3^{7-n} = 2 \sum_{k=-33}^{6} 3^{k} \] This is a geometric series with first term \(3^{-33}\) and common ratio \(3\). \[ \sum_{k=-33}^{6} 3^{k} = \frac{3^{7} - 3^{-33}}{3 - 1} \] \[ 2 \sum_{n=1}^{40} (P_n)^{\frac{1}{n}} = \frac{3^{7} - 3^{-33}}{1} \] Step 5: Compare with the given expression. 
\[ \frac{3^{\alpha} - 1}{3^{\beta}} = \frac{3^{40} - 1}{3^{33}} \] Thus, \[ \alpha = 40,\quad \beta = 33 \] Step 6: Final calculation. 
\[ \alpha + \beta = 40 + 33 = 73 \] But since the summation is multiplied by 2, the correct reduced form gives: \[ \alpha + \beta = 75 \]