如果有理数a和b满足等式|a×b-2|+|1-b|=0,试求1/a×b+1/(a+1)(b+1)+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2011)的值.求洗讲
问题描述:
如果有理数a和b满足等式|a×b-2|+|1-b|=0,试求1/a×b+1/(a+1)(b+1)+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2011)的值.求洗讲
求细讲。
答
有理数a,b满足|ab-2|+|1-b|=0ab-2=0,1-b=0解得a=2,b=11/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2011)(b+2011)=1/1*2+1/2*3+1/3*4+...+1/2013*2012=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2012-1/2013)=1-1/2013=2012/20...