已知ab是有理数,且满足(a-1)²+丨b-2丨=0,求ab分之1+ (a+1)(b+1)分之1+(a+2)(b+2)分之1+····+(a+2012)(b+2012)分之1的值.

问题描述:

已知ab是有理数,且满足(a-1)²+丨b-2丨=0,求ab分之1+ (a+1)(b+1)分之1+(a+2)(b+2)分之1+····+(a+2012)(b+2012)分之1的值.

即a-1=b-2=0
a=1,b=2
所以原式=1/1*2+1/2*3+……+1/2013*2014
=1-1/2+1/2-1/3+……+1/2013-1/2014
=1-2014
=2013/2014

a-1=0
b-2=0
∴a=1
b=2
ab分之1+ (a+1)(b+1)分之1+(a+2)(b+2)分之1+····+(a+2012)(b+2012)分之1
=1/1×2+1/2×3+1/3×4+……+1/2013×2014
=1-1/2+1/2-1/3+1/3-1/4+……+1/2013-1/2014
=1-1/2014
=2013/2014