The term independent of x  in the expansion of (ax+bx)14  is

Given  expansion is (ax+bx)14  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= 14Cr(ax)14−r(bx)rTr+1= 14Cr(a)14−r(x)14−r(b)r(x−1)r (∴xaxb=xa−b)    Tr+1=  14Cra14−rbrx14−2r............(1)x14−2r  compare with x0 because of independent term⇒x14−2r=x0This will be independent of x  if 14−2r=0   ∵r=7 this value substitute in Eqn (1)∴  The term without x  is equal toT8=  14C7a7b7=14!7!7! a7 b7