<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: For a GP with first term a and common ratio r, the product of first n terms is P_n = a^n · r^(n(n-1)/2). Taking the nth root gives P_n^(1/n) = a · r^((n-1)/2), which forms an arithmetic-geometric series that can be summed using calculus or algebraic manipulation.
<p><strong>Step 1: Find P_n (product of first n terms)</strong></p><p>For GP with a₁ = 16, r = 1/4:</p><p>P_n = a₁ · a₂ · a₃ · ... · aₙ = a^n · r^(0+1+2+...+(n-1)) = 16^n · (1/4)^(n(n-1)/2)</p><p><strong>Step 2: Simplify P_n^(1/n)</strong></p><p>P_n^(1/n) = 16 · (1/4)^((n-1)/2) = 16 · (1/2)^(n-1) = 16 · 2^(-(n-1))</p><p>= 16 · 2^(-n+1) = 16 · 2 · 2^(-n) = 32 · (1/2)^n</p><p><strong>Step 3: Sum the infinite series</strong></p><p>∑(n=1 to ∞) P_n^(1/n) = ∑(n=1 to ∞) 32 · (1/2)^n</p><p>= 32 · [1/2 + 1/4 + 1/8 + ...]</p><p>= 32 · (1/2)/(1 - 1/2) = 32 · (1/2)/(1/2) = 32</p><p>∴ Answer: <strong>32</strong></p>
Correct Answer: C