If in the expansion of (x4−1x3)15,x−17  occurs in the rth  term, then

Given expansion  (x4−1x3)15 General term in the expansion(∴  Tr+1= nCr xn−r (a)r  be the expansion of (x+a)n) rth  termTr= T(r−1)+1=15Cr−1(x4)15−r+1(−1x3)r−1             = 15Cr−1x64−4r(−1)r−1x3r−3 (∴xaxb=xa−b)              = 15Cr−1(−1)rx64  −7r+3 This  contains x−17  if x64  −7r+3=x−17 ⇒67−7r=−17 i.e. if 7r  =84  ∵r=12