Question

如果有理数a,b满足ab-3的绝对值+(1-b)平方=0,试求1/ab+1/(a+2)(b+2)+1/ (a+4)(b+4)+...+1/(a+98)(b+98)↓

↑ 的值

Answer 1

解：|ab-3|+(1-b)²=0， 两个非负数的和等于0，这两个非负数都等于0
ab-3=0 且 1-b=0 ，解得 a=3 , b=1

1/ab+1/(a+2)(b+2)+1/(a+4)(b+4)+……+1/(a+98)(b+98)
=1/(1×3) +1/(3×5) +1/(5×7)+……+1/(99×101)
=1/2×[2/(1×3)+2/(3×5)+2/(5×7)+……+2/(99×101) ]
=1/2×[(1-1/3) +(1/3-1/5) +(1/5-1/7)+……+(1/99-1/101) ]
=1/2×(1 -1/3 +1/3-1/5 +1/5-1/7+……+1/99-1/101)
=1/2×( 1 -1/101 )
=1/2×100/101
=50/101