@Nishant Sir : I got the logic Thanks:)
If P = n(n2 - 12)(n2 - 22)(n2 - 32)...(n2 - r2), n > r, n N, then P is divisible by
1. (2r + 2)!
2. (2r - 1)!
3.(2r + 1)!
4.(2r - 2)!
-
UP 0 DOWN 0 0 6
6 Answers
Lokesh Verma
·2009-11-05 20:00:48
see it is the product of 2r+1 consecutive numbers...
hence it is divisible by (2r+1)!
also, because it is divisible by (2r+1)! so it has to be divisible by the smaller factorials
hence the answer should be 2, 3, 4
fibonacci
·2009-11-05 20:07:05
Nishant sir, how are these numbers consecutive.
i think the question should be P=n(n-1)(n-2)...(n-r2)
Lokesh Verma
·2009-11-05 20:12:22
@figonacci
P=n(n-12)(n-22)...(n-r2)
= (n-r)(n-r+1)..........n(n+1)(n+2)......(n+r)