(1+2/3)×(1- 2/3)×(1+2/5)×(1- 2/5)×...×(1+2/99)×(1- 2/99)=?

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(1+2/3)×(1-2/3)×(1+2/5)×(1-2/5)×...×(1+2/99)×(1-2/99)=?(1+2/3)×(1-2/3)×(1+2/5)×(1-2/5)×...×(1+2/99)×

(1+2/3)×(1- 2/3)×(1+2/5)×(1- 2/5)×...×(1+2/99)×(1- 2/99)=?
(1+2/3)×(1- 2/3)×(1+2/5)×(1- 2/5)×...×(1+2/99)×(1- 2/99)=?

(1+2/3)×(1- 2/3)×(1+2/5)×(1- 2/5)×...×(1+2/99)×(1- 2/99)=?
(1+2/3)×(1- 2/3)×(1+2/5)×(1- 2/5)×...×(1+2/99)×(1- 2/99)=[(5/3)*(7/5)*(9/7)*...*(101/99)]*[(1/3)*(3/5)*(5/7)*...(97/99)]=(101/3)*(1/99)=101/297

把题目括号中的数字算出来,找规律
原式=(5/3)*(1/3)*(7/5)*(3/5)*(9/7)*(5/7)*……*(101/99)*(97/99)
每相隔两项之积为1,到最后就剩下了:
(1/3)×(101/99)
=101/297