191问答库 > 1⼀(1^2+1)+1⼀(2^2+2)+1⼀(3^2+3)+...+1⼀(200^2+200)

1⼀(1^2+1)+1⼀(2^2+2)+1⼀(3^2+3)+...+1⼀(200^2+200)

初一的数学题，求简单运算方法。谢谢！

2025-02-23 22:39:34

推荐回答（3个）

回答1：

1//(1^2+1)+1/(2^2+2)+1/(3^2+3)+...+1/(200^2+200)
=1/1(1+1)+1/2(2+1)+1/3(3+1)+..+1/200(200+1)
=1/1*2+1/2*3+1/3*4+....+1/200*201
=1/1-1/2+1/2-1/3+1/3-1/4+....+1/200-1/201
=1-1/201
=200/201

回答2：

1/(1^2+1)+1/(2^2+2)+1/(3^2+3)+...+1/(200^2+200)
=1/(1*2) +1/(2*3)+1/(3*4)……+1/(200*201)
=1/1-1/2+1/2-1/3+1/3+1/4……+1/200-1/201
=1-1/201
=200/201

回答3：

1/(1^2+1)=1/(1*2)=1/1-1/2
1/(2^2+2)=1/(2*3)=1/2-1/3
1/(3^2+3)=1/(3*4)=1/3-1/4
……
1/(1^2+1)+1/(2^2+2)+1/(3^2+3)+...+1/(200^2+200)
=1/1-1/2+1/2-1/3+1/3-1/4+……+1/200-1/201
=1-1/201
=200/201