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)50is(3n+1)51C3 for some positive integer n. Then, the value of n is
5
Coefficient of x2 in the expansion of
{(1+x)2+(1+x)3+⋯+(1+x)49+(1+mx)50}⇒ 2C2+3C2+4C2+⋯+49C2+50C2.m2=(3n+1).51C3⇒ 50C3+50C2m2=(3n+1).51C3[∴rCr+r+1Cr+⋯+nCr=n+1Cr+1]
⇒ 50×49×483×2×1+50×492×m2=(3n+1)51×50×493×2×1⇒ m2=51n+1
∴ Minimum value of m2 for which (51n + 1) is integer (perfect square) for n = 5.
∴ m2=51×5+1⇒m2=256
∴ m = 16 and n = 5
Hence, the value of n is 5.