<p>Find the sum \(\displaystyle\sum_{j=1}^{n}\sum_{i=1}^{n} i \times 3^j\).</p>
Step-by-Step Solution
Key Concept: Separate the double sum into independent variables: ∑∑ i×3^j = (∑i)(∑3^j). Then use standard formulas: ∑i = n(n+1)/2 and ∑3^j = 3(3^n-1)/2.
<p><strong>Step 1:</strong> Recognize that the double sum can be separated since i and j are independent:</p><p>∑(j=1 to n)∑(i=1 to n) i·3^j = ∑(j=1 to n) 3^j · ∑(i=1 to n) i</p><p><strong>Step 2:</strong> Evaluate ∑(i=1 to n) i = n(n+1)/2</p><p><strong>Step 3:</strong> Evaluate ∑(j=1 to n) 3^j using geometric series formula:</p><p>∑(j=1 to n) 3^j = 3 + 3² + ... + 3^n = 3(3^n - 1)/(3 - 1) = 3(3^n - 1)/2</p><p><strong>Step 4:</strong> Multiply the results:</p><p>[n(n+1)/2] × [3(3^n - 1)/2] = 3n(3^n - 1)(n+1)/4</p><p>∴ Answer: <strong>3n(3^n - 1)(n+1)/4</strong></p>
Correct Answer: 3n(3^n - 1)(n+1)/4