(1+1/2)(1+1/3)(1+1/4)...(1+1/2009)(1/2—1)(1/3-1)(1/4-1)…(1/2010-1).(1+1/2)(1+1/3)(1+1/4)…(1+1/2009)(1/2-1)(1/3-1)(1/4-1)…(1/2010-1)这道计算题怎么做呢?~~

问题描述:

(1+1/2)(1+1/3)(1+1/4)...(1+1/2009)(1/2—1)(1/3-1)(1/4-1)…(1/2010-1).
(1+1/2)(1+1/3)(1+1/4)…(1+1/2009)(1/2-1)(1/3-1)(1/4-1)…(1/2010-1)这道计算题怎么做呢?~~

cosA=(b^2+c^2-a^2)/(2bc)=(5-a^2)/4
000-5111

1-1/2+1/2-1/3+1/3-1/4+···+1/2009-1/2010=1+(-1/2+1/2)+(-1/3+1/3)+···--1/2010括号内相加得0则=1-1/2010=2009/2010即(1-1/2)+(1/2-1/3)+(1/3-1/4)+···+(1/2009-1/2010)=2009/2010