Sequences & Series
Summation
Grade 11
Question:
<p>If \(1^2 + 2^2 + 3^2 + \cdots + 2003^2 = (2003)(4007)(334)\) and \((1)(2003) + (2)(2002) + (3)(2001) + \cdots + (2003)(1) = (2003)(334)(x)\), then \(x\) equals</p>
<p>2005</p>
<p>2004</p>
<p>2003</p>
<p>2001</p>
Step-by-Step Solution
Key Concept: Recognize that the sum (1)(2003) + (2)(2002) + (3)(2001) + ... + (2003)(1) can be rewritten as Σk(2004-k) = 2004·Σk - Σk². Use the given formula for Σk² to find x.
<p><strong>Step 1:</strong> Express the second sum algebraically. The general term is k(2004-k) where k runs from 1 to 2003.</p><p>∑[k=1 to 2003] k(2004-k) = ∑[k=1 to 2003] (2004k - k²) = 2004·∑k - ∑k²</p><p><strong>Step 2:</strong> Calculate ∑k from 1 to 2003 using the standard formula.</p><p>∑k = 2003·2004/2 = 2003·1002</p><p><strong>Step 3:</strong> Substitute the known values.</p><p>∑[k=1 to 2003] k(2004-k) = 2004·(2003·1002) - (2003)(4007)(334)</p><p>= 2003[2004·1002 - 4007·334]</p><p>= 2003[2008008 - 1338338]</p><p>= 2003·669670</p><p><strong>Step 4:</strong> Factor 669670 to match the form (2003)(334)(x).</p><p>669670 = 334·2005</p><p>Therefore: ∑k(2004-k) = (2003)(334)(2005)</p><p>∴ x = <strong>2005</strong></p>
Correct Answer: A