Let the sum of given series be S
S=1+(1+b)r+(1+b+b2)r2+(1+b+b2+b3)r3.....
Multiplying both sides by br, we get
Sbr=0+br+(b2+b)r2+(b+b2+b3)r3+(1+b+b2+b3)br4..................
Subtracting both equations, we get
S(1−br)=1+r+r2+r3.............................∞
⇒S(1−br)=11−r
⇒S=1(1−r)(1−br)