a 1b 1+a2 B2+a3 B3+…+anbn = a 1(b 1-B2)+L2(B2-B3)+L3(B3-B4)+…+Ln- 1(bn- 1-bn)+Lnbn、
When n=2, then a1b1+a2b2 = a1(b1-b2)+l2b2,
∴L2=a 1+a2,
When n=3, then a1b1+a2b2+a3b3 = a1(b1-b2)+l2 (b2-b3)+l3b3,
∴L3=a 1+a2+a3,
Generalize this conclusion, LN = A 1+A2+A3+…+An.
So the answer is: a 1+a2+a3, A 1+A2+A3+…+An.