已知丨ab-2丨+丨a-1丨=0,求ab分之1+(a+1)(b+1)分之一+(a+2)(b+2)分之一…+(a+2011
问题描述:
已知丨ab-2丨+丨a-1丨=0,求ab分之1+(a+1)(b+1)分之一+(a+2)(b+2)分之一…+(a+2011
已知丨ab-2丨+丨a-1丨=0,求ab分之1+(a+1)(b+1)分之一+(a+2)(b+2)分之一…+(a+2011)(b+2011)的值。
答
绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=0
ab-2=0
a-1=0
解得a=1 b=2
b=a+1
1/(ab)+1/[(a+1)(b+1)]+1/[(a+2)(b+2)]+...+1/[(a+2011)(b+2011)]
=1/[a(a+1)]+1/[(a+1)(a+2)]+1/[(a+2)(a+3)]+...+1/[(a+2011)(a+2012)]
=1/a -1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3)+...+1/(a+2011)-1/(a+2012)
=1/a -1/(a+2012)
=1/1 -1/(1+2012)
=1 -1/2013
=2012/2013