Question

Let $m$ be the smallest positive integer such that the coefficient of $x^2$ in the expansion of $(1+x{)}^2+(1+x{)}^3+...+(1+x{)}^{49}+(1+mx{)}^{50}\text{is}(3n+1{)}^{51}{C}_3$ for some positive interger $n$. Then the value of $n$ is ___

Answer 1

$(1+x{)}^2+(1+x{)}^3+...+(1+x{)}^{49}+(1+mx{)}^{50}=(1+x{)}^2\left[\frac{(1+x{)}^{48}}{(1+x)-1}\right]+(1+mx{)}^{50}=\frac{1}{x}\left[\right(1+x{)}^{50}-(1+x{)}^2)]+(1+mx{)}^{50}\text{coeff. of}{x}^2\text{in the above expansion}=\text{coeff. of}{x}^3\text{in}(1+x{)}^{50}+\text{Coeff. of}{x}^2\text{in}(1+mx{)}^{50}\Rightarrow {}^{50}{C}_3+{}^{50}{C}_2{m}^2\therefore (3n+1{)}^{51}{C}_3={}^{50}{C}_3+{}^{50}{C}_2{m}^2\Rightarrow (3n+1)=\frac{{}^{50}{C}_3}{{}^{51}{C}_3}+\frac{{}^{50}{C}_2}{{}^{51}{C}_3}{m}^2\Rightarrow 3n+1=\frac{16}{17}+\frac{1}{17}{m}^2\Rightarrow n=\frac{m^2-1}{51}$
Least positive integer $m$ for which $n$ is an interger is $m=16$ and then $n=5$