Question

(1*1+2*2)/(1*2)+(2*2+3*3)/(2*3)+(3*3+4*4)/(3*4)...+(2000*2000+2001*2001)/(2000*2001)=?

Answer 1

这样的题一般是要对每一项进行拆分
抽象出来应该是:
(n^2+(n+1)^2)/(n*(n+1))
=n/(n+1) + (n+1)/n
=1-1/(n+1) + 1 + 1/n
=2 + 1/n - 1/(n+1)
所以原式= 2*2000 + 1/1-1/2 + 1/2 -1/3 + ……+ 1/2000 - 1/2001
=4000+1-1/2001
化简略

Answer 2

(1*1+2*2)/(1*2)
=(1*1+2*2)/(1*2) - 2+2
=(1*1+2*2 - 2 * 1 * 2)/(1*2)
=(2-1)^2 / (1*2) = 1^2 / (1*2) = 1/1 - 1/2

所以：(1*1+2*2)/(1*2)+(2*2+3*3)/(2*3)+(3*3+4*4)/(3*4)...+(2000*2000+2001*2001)/(2000*2001)=(1*1+2*2 - 2*1*2)/(1*2)+(2*2+3*3-2*2*3)/(2*3)+(3*3+4*4-2*3*4)/(3*4)...+(2000*2000+2001*2001-2*2000*2001)/(2000*2001) + 2000*2
=(1)/(1*2)+(1)/(2*3)+(1)/(3*4)...+(1)/(2000*2001) + 4000
=1-1/2 + 1/2-1/3 +...+1/2000-1/2001 +4000
=4001-1/2001
=4000+2000/2001
=8006000/2001

Answer 3

(1*1+2*2)/(1*2)+(2*2+3*3)/(2*3)+(3*3+4*4)/(3*4)...+(2000*2000+2001*2001)/(2000*2001)
=[2+1/(1*2)]+[2+1/(2*3)]+[2+1/(3*4)]+...+[2+1/(2000*2001)]
=2*2000+1/(1*2)+1/(2*3)+1/(3*4)+...+1/(2000*2001)
=4000+(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2000-1/2001)
=4000+(1-1/2+1/2-1/3+1/3-1/4+...+1/2000-1/2001)
=4000+(1-1/2001)
=4000+2000/2001
=4000又2000/2001
或8006000/2001
或4000.9995002498750624687656171914......

Answer 4

4000 2000/2001

Answer 5

手机照的不是很清楚，不过能看出来。