UPSC NDA II 2024 Mathematics Paper

Concept:The problem uses the binomial expansion of $(6+1)^n$ to show that all terms except the constant term are multiples of 36.Explanation:Write $7^n - 6n$ as $(6+1)^n - 6n$.Expand $(6+1)^n$ using the binomial theorem:$(6+1)^n = 1 + 6n + \binom{n}{2}6^2 + \binom{n}{3}6^3 + \cdots + 6^n$.Subtract $6n$:$(1 + 6n + \binom{n}{2}6^2 + \cdots + 6^n) - 6n = 1 + \binom{n}{2}6^2 + \binom{n}{3}6^3 + \cdots + 6^n$.Factor $6^2 = 36$ from the remaining terms (except the constant 1):$= 1 + 36\left(\binom{n}{2} + \binom{n}{3}6 + \cdots + 6^{n-2}\right)$.For $n=100$, the expression becomes $1 + 36k$, where $k = \binom{100}{2} + \binom{100}{3}\cdot 6 + \cdots + 6^{98}$ is an integer.Thus, when divided by 36, the remainder is 1.Answer:Option B: 1.