Q.

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]


Ans.

(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