If the expansion in powers of x of the function 1/[(1−ax)(1−bx)] is a0+a1x+a2x2+a3x3+⋯, then an is

1(1−ax)(1−bx)=a0+a1x+a2x2+⋯+anxn+⋯But (1−ax)−1(1−bx)−1=1+ax+a2x2+⋯×1+bx+b2x2+⋯⇒  Coefficient of xn is bn+abn−1+a2bn−2+⋯+an−1b+an=bn+1−an+1b−a⇒ an=bn+1−an+1b−a