Question

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

Answer 1

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