Let  n be the smallest positive integer such that the coefficients  x2 in the expansion (1+x)2+(1+x)3+…+(1+x)49+(1+mx)50  is  (3n+1)51C3 for some positive integer n. Then the value of n , is

We find that  (1+x)2+(1+x)3+…(1+1)40+(1+mx)50=(1+x)2(1+x)48−1(1+x)−1+(1+mx)30=1x(1+x)50−(1+x)2+(1+mx)30∴ Coeffi. of x2 in (1+x)2+(1+x)3+…(1+x)49+(1+mx)50= Coeffi. of x2 in 1x(1+x)30−(1+x)2+(1+mx)50= Coeffi. of x3 in (1+x)50−(1+x)2+ coefficient of x2 in (1+mx)50=50C3+51C2m2It is given that this coefficient is (3n+1)51C3 ∴ 50C3+50C2m2=(3n+1)51C3⇒ 50×49×483×2×1+50×492×1m2=(3n+1)51×50×493×2×1⇒ 16+m2=17(3n+1)⇒ m2=51n+1The least value of n for which  51n+1 is a perfect square is 5 and for this value of  n the value of  m is 16. Hence, m=16 and  n =5.