<p>How many terms of \(1 + 3 + 5 + 7 + \ldots\) amount to 1234321?</p>
Step-by-Step Solution
Key Concept: Recognize that 1 + 3 + 5 + ... + (2n-1) = n². Therefore, find n such that n² = 1234321, which means n = √1234321.
<p><strong>Step 1:</strong> Identify the series pattern. This is the sum of the first n odd numbers: 1 + 3 + 5 + 7 + ... + (2n-1)</p><p><strong>Step 2:</strong> Recall the key property: The sum of the first n odd natural numbers equals n². That is, 1 + 3 + 5 + ... + (2n-1) = n²</p><p><strong>Step 3:</strong> Set up the equation: n² = 1234321</p><p><strong>Step 4:</strong> Solve for n by taking the square root: n = √1234321 = 1111</p><p><strong>Step 5:</strong> Verify: 1111² = 1234321 ✓</p><p><strong>Verification:</strong> The last term is 2(1111) - 1 = 2221, and indeed summing odd numbers from 1 to 2221 gives 1111² = 1234321</p><p>∴ Answer: <strong>1111 terms</strong></p>
Correct Answer: 1111