Let m be the smallest positive integer such that the coefficient of x2 in the expansion of (1+x)2+(1+x)3+…..+(1+x)49+(1+mx)50 is  (3n+1)·51C3for some positive integer n. Then the value of n is

(1+x)2+(1+x)3+............+(1+x)49+(1+mx)50=(1+x)2(1+x)48−11+x−1+(1+mx)50=(1+x)50−(1+x)2x+(1+mx)50Coeff  of   x2   in   above=50C3+50C2m2=(3n+1) 51C3⇒50×49×486+50×492m2=(3n+1)51×50×496⇒8+m22=(3n+1)172⇒16+m2=51n+17⇒m2=51n+1m,n∈z+;⇒n=5,  m=16