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) C351 for some positive integer n. Then the value of n is [2016]
(5)
(1+x)2+(1+x)3+⋯+(1+x)49+(1+mx)50
=(1+x)2[(1+x)48-1(1+x)-1]+(1+mx)50
=1x[(1+x)50-(1+x)2]+(1+mx)50
Coeff. of x2 in the above expansion
=Coeff. of x3 in (1+x)50+Coeff. of x2 in (1+mx)50
=C350+C250m2
∴ (3n+1) C351=C350+C250m2
⇒(3n+1)=C350C351+C250C351m2
⇒3n+1=1617+117m2
⇒n=m2-151
∴ Least positive integer m for which n is an integer is m=16 and then n=5