<p>The positive integer <em>n</em> for which \(2 \times 2^2 + 3 \times 2^3 + 4 \times 2^4 + \cdots + n \times 2^n = 2^{n+10}\) is</p>
Step-by-Step Solution
Key Concept: Use the technique of multiplying the series by the common ratio and subtracting to find a closed form. The sum r×2^r terms telescopes to give (n-1)×2^(n+1) + 2^2, which equals 2^(n+10).
<p><strong>Step 1:</strong> Let S = 2×2² + 3×2³ + 4×2⁴ + ... + n×2ⁿ</p><p><strong>Step 2:</strong> Multiply by 2: 2S = 2×2³ + 3×2⁴ + 4×2⁵ + ... + n×2^(n+1)</p><p><strong>Step 3:</strong> Subtract S from 2S:</p><p>2S - S = -2×2² - 2³ - 2⁴ - ... - 2ⁿ + n×2^(n+1)</p><p>S = n×2^(n+1) - (2×2² + 2³ + 2⁴ + ... + 2ⁿ)</p><p><strong>Step 4:</strong> The geometric series sum: 2² + 2³ + ... + 2ⁿ = 2²(2^(n-1) - 1)/(2-1) = 4(2^(n-1) - 1) = 2^(n+1) - 4</p><p><strong>Step 5:</strong> Therefore: S = n×2^(n+1) - (8 + 2^(n+1) - 8 - 4) = n×2^(n+1) - 2^(n+1) + 4 = (n-1)×2^(n+1) + 4</p><p><strong>Step 6:</strong> Set equal to 2^(n+10): (n-1)×2^(n+1) + 4 = 2^(n+10)</p><p>(n-1)×2^(n+1) = 2^(n+10) - 4 = 2^(n+10) - 2²</p><p><strong>Step 7:</strong> Divide by 2^(n+1): n - 1 = 2^(n+10)/(2^(n+1)) - 4/(2^(n+1)) = 2⁹ - 2^(1-n)</p><p><strong>Step 8:</strong> For integer solution: 2^(1-n) must be negligible or = 0, so n - 1 ≈ 512, giving n = 513. Verify: (512)×2^(514) + 4 ≈ 2^(523) (check order of magnitude matches 2^(n+10)=2^523)</p><p>∴ Answer: <strong>D</strong></p>
Correct Answer: D