The coefficient of x−3  in the expansion of (x−mx)11  is

Given expansion (x−mx)11  is We have general term in the expansion (x+a)n (∴  Tr+1= nCr xn−r (a)r  be the expansion of (x+a)n) Tr+1=11Cr x11−r (−mx)r Tr+1=11Cr x11−r (−m)r (x−1)r Tr+1= 11Cr (−1)r m2 x11−2r..........................(1) x11−2r compare with x−3  for finding r ⇒x11−2r=x−3  And this contains x−3  only if 11−2r=−3  ∵r=7 this value substitute in Eqn (1) T7+1= 11C7 (−1)7 m2 x11−2×7 T8= −11C7 m2 x−3 So, T8  contains x−3  and             T8= 11C7(−1)7m7x−3                   =  11.10.9.81.2.3.4(−1)7m7  =−330 m7