<p>Find the sum of the series \(1^2 + 3^2 + 5^2 + \cdots\) to <em>n</em> terms.</p>
Step-by-Step Solution
Key Concept: Recognize this as the sum of squares of the first n odd numbers. Use the formula for sum of squares of first (2n-1) consecutive integers, then subtract the sum of squares of even numbers.
<p><strong>Step 1:</strong> Identify the series. The nth odd number is (2n-1), so we need: 1² + 3² + 5² + ... + (2n-1)²</p><p><strong>Step 2:</strong> Use the identity that sum of squares of first m integers is m(m+1)(2m+1)/6. Here we need the sum up to (2n-1), so m = 2n-1:</p><p>Sum of 1² + 2² + 3² + ... + (2n-1)² = (2n-1)(2n)(4n-1)/6</p><p><strong>Step 3:</strong> Subtract the sum of squares of even numbers (2² + 4² + 6² + ... + (2n-2)²):</p><p>Even sum = 4(1² + 2² + 3² + ... + (n-1)²) = 4·(n-1)(n)(2n-1)/6 = 2(n-1)(n)(2n-1)/3</p><p><strong>Step 4:</strong> Therefore: Odd sum = (2n-1)(2n)(4n-1)/6 - 2(n-1)(n)(2n-1)/3</p><p>= (2n-1)/6 [2n(4n-1) - 4(n-1)n]</p><p>= (2n-1)/6 [8n² - 2n - 4n² + 4n]</p><p>= (2n-1)/6 [4n² + 2n]</p><p>= (2n-1)·2n(2n+1)/6</p><p>∴ Answer: <strong>n(2n-1)(2n+1)/3</strong></p>
Correct Answer: n(2n-1)(2n+1)/3