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) 51C3 for some positive integer n. Then the value of n is _________.

Coefficient of x2 in expansion=1+3C2+4C2+5C2+⋯+49C2+50C2⋅m2       as nCr+nCr−1=n+1Cr= 3C3+3C2+4C2+5C2+⋯+49C2+50C2⋅m2= 4C3+4C2+⋯+50C2m2=5C3+⋯+49C2+50C2m2=50C3+50C2m2+50C2−50C2=51C3+50C2m2−1=(3n+1) 51C3( given )∴ 3n⋅51350C2=50C2m2−1m2−151=n⇒m2-1=51n⇒m2=51n+1n=5⇒m2=256value of n is 5.