Ta có: an+bn=(a+b)(an−1−an−2b+an−3.b2−...+a2bn−3−a.bn−2+bn−1)∀n∈N∗ và n lẽ
Nên: (an+bn)⋮(a+b)∀n∈N∗ , n lẽ 11991+21991+31991+...+19911991=11991+21991+31991+...+19901991+19911991=[(11991+19901991)+(21991+19891991)+...+(9951991+9961991)+19911991]⋮1991
Hay: .... chia hết cho 11