如果有理数a,b满足丨ab-2丨+(1-b)^2=0 试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)
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如果有理数a,b满足丨ab-2丨+(1-b)^2=0 试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2004)(b+2004)的值?
首先可以看出:|ab-2|=-(1-b)^2
很明显-> b=1;a=2:
然后带入运算:->
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2) +……+1/(a+2004)(b+2004)
=1/2+1/(3*2) + 1/(3*4) + 1/(4*5) + .
+1/(2005*2006)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.
+(1/2005-1/2006)
注释:(化简后可各项相销)
=1-1/2006
=2005/2006
这便是最终结果了.
很明显-> b=1;a=2:
然后带入运算:->
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2) +……+1/(a+2004)(b+2004)
=1/2+1/(3*2) + 1/(3*4) + 1/(4*5) + .
+1/(2005*2006)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.
+(1/2005-1/2006)
注释:(化简后可各项相销)
=1-1/2006
=2005/2006
这便是最终结果了.
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