已知Iab-2I和Ib-1I互为相反数,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2009)(b+2009)
问题描述:
已知Iab-2I和Ib-1I互为相反数,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2009)(b+2009)
的值
答
已知Iab-2I和Ib-1I互为相反数,则Iab-2I=0,Ib-1I=0,所以a=2,b=1.
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2009)(b+2009)
=1/2 + 1 / (2*3) + 1/(3*4) + ……+ 1/(2010*2011)
=1- 1/2 + 1/2 -1/3 + 1/3- 1/4 + ……+ 1/2010 - 1/2011
=1-1/2011
=2010/2011