Question

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

Answer 1

Answer: 5

$(1+x{)}^2+(1+x{)}^3+\dots +(1+x{)}^{49}+(1+mx{)}^{50}$
$=\frac{(1+x{)}^2\left[(1+x{)}^{48}-1\right]}{(1+x)-1}+(1+mx{)}^{50}$
$=\frac{(1+x{)}^{50}}{x}-\frac{(1+x{)}^2}{x}+(1+mx{)}^{50}$
given that coefficient of $x^2=(3n+1){}^{51}{C}_3$
$\therefore {}^{50}{C}_3+{}^{50}{C}_2{m}^2=(3n+1){}^{51}{C}_3$
after simplifying
$\Rightarrow \frac{{}^{50}{C}_3}{{}^{50}{C}_2}+m^2=(3n+1)\frac{{}^{51}{C}_3}{{}^{50}{C}_2}$
$\Rightarrow \frac{50-2}{3}+m^2=(3n+1)\frac{51}{3}$
$\Rightarrow 16+m^2=17(3n+1)$
$\Rightarrow m^2=51n+1$
$m$ is the smallest +ve integer
$\therefore n=5$