Let m be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of $(1 + x)^{2} + (1 + x)^{3} + \ldots + (1 + x)^{49} + (1 + mx)^{50}$ is $(3n + 1)\,{}^{51}C_{3}$ for some positive integer n. Then the value of n is

$\text{Given: } S = (1+x)^{2} + (1+x)^{3} + \ldots + (1+x)^{49} + (1+mx)^{50}$

$[x^{2}]\,S = (3n+1)\,{}^{51}C_{3}$

$\text{To find: the value of } n$

$(1+x)^{k} = \ldots + {}^{k}C_{2}\,x^{2} + \ldots$ $[x^{2}]\,(1+x)^{k} = {}^{k}C_{2}$ $\sum_{k=2}^{49} {}^{k}C_{2} = {}^{50}C_{3}$ $[x^{2}]\,(1+mx)^{50} = {}^{50}C_{2}\,m^{2}$ $[x^{2}]\,S = {}^{50}C_{3} + {}^{50}C_{2}\,m^{2} = (3n+1)\,{}^{51}C_{3}$ ${}^{50}C_{3} = 19600, \quad {}^{50}C_{2} = 1225, \quad {}^{51}C_{3} = 20825$ $19600 = 16 \cdot 1225, \qquad 20825 = 17 \cdot 1225$ $1225\,(16 + m^{2}) = 1225 \cdot 17\,(3n+1)$ $\Rightarrow 16 + m^{2} = 17(3n+1)$ $m^{2} = 51n + 1 \ \Rightarrow \ m^{2} \equiv 1 \pmod{51}$ $51 = 3 \cdot 17; \quad m \equiv \pm 1 \ (\text{mod } 3), \ m \equiv \pm 1 \ (\text{mod } 17)$ $m \equiv 1, 16, 35, 50 \ (\text{mod } 51) \ \Rightarrow \ m_{\min} = 16$ $m = 16 \ \Rightarrow \ n = \dfrac{m^{2}-1}{51} = \dfrac{256-1}{51} = \dfrac{255}{51} = 5$ $1225 \cdot 272 = 333200 = 16 \cdot 20825$ $\text{both sides match}$ $\boxed{n = 5}$