230问答网 > 计算(1*2分之1的2次方+2的2次方)+(2*3分之2的2次方+3的2次方)+(3*4分之3的2次方+4的2次方)+……+

计算(12分之1的2次方+2的2次方)+(23分之2的2次方+3的2次方)+(3*4分之3的2次方+4的2次方)+……+

2024-12-16 14:27:40

推荐回答（1个）

回答1：

考察一般项：
[n²+(n+1)²]/[n(n+1)]
=(2n²+2n+1)/[n(n+1)]
=[2n(n+1)+1]/[n(n+1)]
=2 +1/[n(n+1)]
=1/n -1/(n+1) +2

(1²+2²)/(1×2)+(2²+3²)/(2×3)+(3²+4²)/(3×4)+...+(2000²+2001²)/(2000×2001)+(2001²+2002²)/(2001×2002)
=(1/1-1/2 +2)+(1/2-1/3 +2)+(1/3-1/4+2)+...+(1/2000-1/2001+2)+(1/2001-1/2002+2)
=(1-1/2+1/2-1/3+1/3-1/4+...+1/2000-1/2001+1/2001-1/2002)+2×2001
=(1-1/2002)+4002
=4002又2001/2002