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求积(1+1/1*3)(1+1/2*4)(1+1/3*5).(1+1/99*101)的整数部分
题目详情
求积(1+1/1*3)(1+1/2*4)(1+1/3*5).(1+1/99*101)的整数部分
▼优质解答
答案和解析
式子1+1/1*3=(2/1)*(2/3),
1+1/2*4=(3/2)*(3/4).
(1+1/1*3)(1+1/2*4)(1+1/3*5).(1+1/99*101)
=[(2/1)*(2/3)]*[(3/2)*(3/4)]*[(4/3)*(4/5)]……[(100/99)*(100/101)]
=2*【(2/3)*(3/2)】*[(3/4)*(4/3)]*[4/5…………（100、99)]*(100/101)
=200/101
=1.98
1+1/2*4=(3/2)*(3/4).
(1+1/1*3)(1+1/2*4)(1+1/3*5).(1+1/99*101)
=[(2/1)*(2/3)]*[(3/2)*(3/4)]*[(4/3)*(4/5)]……[(100/99)*(100/101)]
=2*【(2/3)*(3/2)】*[(3/4)*(4/3)]*[4/5…………（100、99)]*(100/101)
=200/101
=1.98
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