In the expansion of (x4−1x3)15 , the coefficient of x39  is

Given expansion  (x4−1x3)15 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= 15Cr(x4)15−r(−1x3)rTr+1= 15Cr (x4)15−r (−1)r (x−3)r  Tr+1= 15Cr x60−4r−3r(−1)rTr+1= 15Cr x60−7r(−1)r.........................(1)x60−7r compare with x39  because of finding r⇒x60−7r=x39This will contain x39  if 60−7r=39∵r=3 this value substitute in Eqn (1)T3+1= 15C3 x60−7×3(−1)3T4= −15C3 x39T4= −15×14×133×2×1  x39∴   T3+1  i.e. T4  is the term containing x39∴  Required coefficient  =  (−1)3  15C3      =−15×14×131×2×3=−455