UPSC CDS II 2024 Math Paper

Concept:The expression $p = n(n+1)(n+2)(n+3) + 1$ represents the product of four consecutive natural numbers plus one, which is always a perfect square and always odd.Explanation:Statement I (p is always odd):Among any four consecutive natural numbers, at least two are even.The product $n(n+1)(n+2)(n+3)$ therefore contains at least one factor of 2, making it even.Adding 1 to an even number gives an odd number.Hence, $p$ is always odd. So Statement I is correct.Statement II (p is a perfect square):Rewrite the product in pairs:$n(n+3) = n^2 + 3n$ and $(n+1)(n+2) = n^2 + 3n + 2$.Let $t = n^2 + 3n$. Then:$p = t(t+2) + 1 = t^2 + 2t + 1 = (t+1)^2$.Since $t+1 = n^2 + 3n + 1$ is an integer, $p$ is a perfect square for every natural number $n$.For example:$n=1$: $p = 1\cdot2\cdot3\cdot4 + 1 = 25 = 5^2$$n=2$: $p = 2\cdot3\cdot4\cdot5 + 1 = 121 = 11^2$$n=3$: $p = 3\cdot4\cdot5\cdot6 + 1 = 361 = 19^2$Thus Statement II is also correct.Answer:Both statements I and II are correct.