(x²+1/x)⁶的展开式中x³的系数为
T‹r+1›=C(6,r)[(x²)^(6-r)][(x⁻¹)^r]=C(6,r)x^(12-3r)
令12-3r=3,得r=3；即第4项的是含x³的项；其系数=C(6,3)=6×5×4/3!=20.