<p>Let \(a_n\) be a sequence in geometric progression with first term 16 and common ratio \(\dfrac{1}{4}\). Let \(P_n\) be the product of first \(n\) terms of the given geometric progression. The value of \(\displaystyle\sum_{n=1}^{\infty} P_n^{1/n}\), is:</p>
Step-by-Step Solution
Key Concept: Each P_n is a product of n terms in GP, so P_n^(1/n) is the geometric mean of these terms. For a GP, this simplifies to a_1 · r^((n-1)/2), and the infinite series becomes a geometric series whose sum can be evaluated.
<p><strong>Step 1: Express the nth term of the GP</strong></p><p>Given: first term a₁ = 16, common ratio r = 1/4</p><p>So aₙ = 16·(1/4)^(n-1) = 16·4^(-(n-1)) = 2^(4)·2^(-2(n-1)) = 2^(6-2n)</p><p><strong>Step 2: Find Pₙ (product of first n terms)</strong></p><p>Pₙ = a₁·a₂·a₃·...·aₙ</p><p>Using the formula for product of n terms in GP: Pₙ = a₁ⁿ·r^(0+1+2+...+(n-1)) = a₁ⁿ·r^(n(n-1)/2)</p><p>Pₙ = 16ⁿ·(1/4)^(n(n-1)/2) = 2^(4n)·2^(-2n(n-1)/2) = 2^(4n-n(n-1)) = 2^(4n-n²+n) = 2^(5n-n²)</p><p><strong>Step 3: Calculate Pₙ^(1/n)</strong></p><p>Pₙ^(1/n) = (2^(5n-n²))^(1/n) = 2^((5n-n²)/n) = 2^(5-n)</p><p><strong>Step 4: Evaluate the infinite series</strong></p><p>∑(n=1 to ∞) Pₙ^(1/n) = ∑(n=1 to ∞) 2^(5-n) = 2⁵·∑(n=1 to ∞) 2^(-n)</p><p>= 32·∑(n=1 to ∞) (1/2)ⁿ = 32·[1/2 + 1/4 + 1/8 + ...]</p><p>= 32·[1/2·1/(1-1/2)] = 32·[1/2·2] = 32·1 = 32</p><p>∴ Answer: <strong>32</strong></p>
Correct Answer: C