KVPY 2020-21 SA Exam Previous Papers

I does not hold for n = 4
⇒ I is false
For statement II
n(n+1)2|n!
⇒(n+1)2|(n−1)!
Let n = 3k – 1, k > 3, k ∈ N
n + 1 = 3k, n – 1 = 3k – 2
(n – 1)! = (3k – 2)!
= (3k – 2) × (3(k – 1)) × … × (2k + 1) (2k) (2k – 1) …× (k + 1) k(k – 1) … 3 × 2 × 1
RHS contains 32,k2 hence is divisible by (3k)2.
⇒ (n – 1)! is divisible by (n+1)2 ⇒ II is true.