For n=1 it's 1, and we claim for all n>=2 it's (x+b1)(x+b2)...(x+bn)/b1.b2....bn, which follows easily by induction.
I.e. Let the sum of n terms be Sn, then S2 = 1 + x/b1 = (x+b1)/b1 which is what we want,
Then if it follows up to some n=k, for n=k+1, we have
S(k+1) = Sk + x(x+b1)(x+b2)...(x+bk)/b1.b2...b(k+1)
= (x+b1)(x+b2)...(x+bk)/b1.b2...bk + x(x+b1)(x+b2)...(x+bk)/b1.b2...b(k+1)
= (1+x/b(k+1))((x+b1)(x+b2)...(x+bk)/b1.b2...bk)
= ((x+b(k+1))/b(k+1))((x+b1)(x+b2)...(x+bk)/b1.b2...bk)
= (x+b1)(x+b2)...(x+b(k+1))/b1.b2...b(k+1) as required
