<p>The greatest integer by which \(1 + \sum_{r=1}^{30} r \times r!\) is divisible is</p>
Step-by-Step Solution
Key Concept: Recognize that r × r! = (r+1)! - r!, which converts the sum into a telescoping series. The sum becomes (31! - 1! ), and you need to find the greatest integer dividing this expression.
<p><strong>Step 1:</strong> Observe that r × r! = (r+1)! - r! (since (r+1)! = (r+1) × r! = r × r! + r!)</p><p><strong>Step 2:</strong> Apply this identity to the sum:<br>∑(r=1 to 30) r × r! = ∑(r=1 to 30) [(r+1)! - r!]</p><p><strong>Step 3:</strong> This is a telescoping series:<br>= (2! - 1!) + (3! - 2!) + (4! - 3!) + ... + (31! - 30!)<br>= 31! - 1!</p><p><strong>Step 4:</strong> Calculate 31! - 1. Since 1! = 1:<br>Sum = 31! - 1</p><p><strong>Step 5:</strong> The expression 1 + (31! - 1) = 31!, and the greatest integer dividing 31! is <strong>31!</strong> itself, or more practically in JEE context, we seek the GCD. Since the sum equals 31! - 1, and these are consecutive integers, gcd(31! - 1, any divisor of 31!) = 1. However, the greatest integer dividing the original expression 1 + ∑ r×r! is <strong>31!</strong>.</p><p>∴ Answer: A</p>
Correct Answer: A