规律:1/n(n+1)=1/n-1/(n+1) (即连续两个自然数的积的倒数,等于这这两个数倒数的差)
裂项相减法:每一项都可以化成两个分数的差
1/1*2+1/2*3+1/3*4+...+1/98*99+1/99*100
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/98-1/99)+(1/99-1/100)
=1-1/2+1/2-1/3+1/3-1/4+……+1/98-1/99+1/99-1/100
=1-1/100
=99/100
规律:1/n(n+1)=1/n-1/(n+1) (即连续两个自然数的积的倒数,等于这这两个数倒数的差)
裂项相减法:每一项都可以化成两个分数的差
1/1*2+1/2*3+1/3*4+...+1/98*99+1/99*100
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/98-1/99)+(1/99-1/100)
=1-1/2+1/2-1/3+1/3-1/4+……+1/98-1/99+1/99-1/100
=1-1/100
=99/100