来自陈立国的问题
计算(a1+a2+…+an-1)(a2+a3+…+an-1+an)-(a2+a3+…+an-1)(a1+a2+…+an)=______.
计算(a1+a2+…+an-1)(a2+a3+…+an-1+an)-(a2+a3+…+an-1)(a1+a2+…+an)=______.
1回答
2020-10-29 16:45
计算(a1+a2+…+an-1)(a2+a3+…+an-1+an)-(a2+a3+…+an-1)(a1+a2+…+an)=______.
计算(a1+a2+…+an-1)(a2+a3+…+an-1+an)-(a2+a3+…+an-1)(a1+a2+…+an)=______.
设x=a1+a2+…+an,y=a2+a3+…+an-1,
则原式=(x-an)(y+an)-yx
=xy+xan-any-an2-xy
=an(x-y)-an2
=an[(a1+a2+…+an)-(a2+a3+…+an-1)]-an2
=an(a1+an)-an2
=a1an,
故答案为:a1an.