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If the sum S = (sum from r=0 to 18 of (r+1)! * (r² + 4(r+1))) + 2 equals k!, find the value of k.
- 20
- 21
- 22
- 23
Correct answer: 21
Solution
We have (r+1)! * (r² + 4(r+1)) = (r+1)! * (r² + 4r + 4) = (r+1)! * (r+2)². Now (r+1)! * (r+2)² = (r+2)! * (r+2) = (r+2)*[(r+3) - 1]*(r+1)! — let us do it cleanly: (r+1)!*(r+2)² = (r+2)!*(r+2) = (r+3)! - (r+2)!. So the sum telescopes: sum from r=0 to 18 of [(r+3)! - (r+2)!] = 21! - 2!. Adding 2: 21! - 2! + 2 = 21! - 2 + 2 = 21!. So k! = 21!, giving k = 21.
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