1)

Let m be the smallesr positive integer such that the coefficient of  $x^{2}$  in the  expansion of $(1+x)^{2}+(1+x)^{3}+.....+(1+x)^{49}+(1+mx)^{50}$   is

   $\left(3n+1\right)^{51}C_{3}$  for some positive integer  n, Then the value of n is


A) 5

B) 7

C) 4

D) 8

Answer:

Option A

Explanation:

Coefficient of $x^{2}$  in the expansion of

  {$(1+x)^{2}+(1+x)^{3}+.....+(1+x)^{49}+(1+mx)^{50}$} 

$\Rightarrow$    $^{2}C_{2}+^{3}C_{2}+^{4}C_{2}+......+^{49}C_{2}+^{50}C_{2}.m^{2}$

$= (3n+1).^{51}C_{3}$

$\Rightarrow$     $^{50}C_{3}+^{50}C_{2}m^{2}=(3n+1).^{51}C_{3}$

[$\because $   $^{r}C_{r}+^{r+1}C_{r}+.....+^{n}C_{r}=^{n+1}C_{r+1}$  ]

$\Rightarrow$     $\frac{50\times 49\times48}{3\times2\times1}+\frac{50\times49}{2}\times m^{2}$

$=(3n+1)\frac{51\times 50\times 49}{3\times2\times1}$

$m^{2}=51n+1$

$\therefore$    Minimum  value of  $m^{2}$   for which (51n+1)  is integer (perfect square) for n=5.

$\therefore$    $m^{2}=51\times5+1$

$\Rightarrow$     $m^{2}=256$

$\therefore$         m=16  and n=5    

Hence, the value of n is 5