KVPY 2020-21 SA Exam Previous Papers

I does not hold for n = 4
⇒ I is false
For statement II
$\mathit{n}{\left(\mathit{n}+1\right)}^2|\mathit{n}!$
$\Rightarrow {\left(\mathit{n}+1\right)}^2|\left(\mathit{n}-1\right)!$
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 $3^2,{\mathit{k}}^2$ hence is divisible by ${\left(3\mathit{k}\right)}^{2.}$
⇒ (n – 1)! is divisible by ${\left(\mathit{n}+1\right)}^2$ ⇒ II is true.