<p>Find the value of \(\displaystyle\sum_{1 \le i < j} i \times \left(\dfrac{1}{2}\right)^j\).</p>
Step-by-Step Solution
Key Concept: Recognize that the sum counts each pair (i,j) where i < j and both divide n. This equals the number of divisor pairs, which can be computed using the divisor function: if n has d(n) divisors, the answer is C(d(n),2) = d(n)(d(n)-1)/2. For n=6 with divisors {1,2,3,6}, we get C(4,2)=6.
<p><strong>Step 1:</strong> Interpret the sum. We need ∑(1≤i<j≤n, i|n, j|n) 1, which counts pairs of divisors (i,j) where i < j.</p><p><strong>Step 2:</strong> Find all divisors of n=6: {1, 2, 3, 6}. Total divisors d(6) = 4.</p><p><strong>Step 3:</strong> The number of ways to choose 2 distinct divisors where order matters (i < j) is C(4,2) = 4·3/2 = 6.</p><p><strong>Step 4:</strong> List the pairs: (1,2), (1,3), (1,6), (2,3), (2,6), (3,6). Each pair contributes 1 to the sum.</p><p>∴ Answer: 6</p>
Correct Answer: 2