若有理数a,b满足/ab-2/+(1-b)的平方=0,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...(1/(a+99)(b+99)的值

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若有理数a,b满足/ab-2/+(1-b)的平方=0,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...(1/(a+99)(b+99)的值

若有理数a,b满足/ab-2/+(1-b)的平方=0,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...(1/(a+99)(b+99)的值
若有理数a,b满足/ab-2/+(1-b)的平方=0,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...(1/(a+99)(b+99)的值

若有理数a,b满足/ab-2/+(1-b)的平方=0,试求:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...(1/(a+99)(b+99)的值
由于/ab-2/+(1-b)的平方=0,所以b=1,a=2
所以
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...(1/(a+99)(b+99)
=1/2+1/(3*2)+1/(4*3)+...+1/(101*100)
=(1-1/2)+(1/2-1/3)+...+(1/100-1/101)
=1-1/101
=100/101