詳解數學 分項對消法： 2/1*4-2/4*7-2/7*10-2/10*13-2/13*16?
回答 (1)
2016-11-26 2:13 am
2/(1*4) - 2/(4*7) - 2/(7*10) - 2/(10*13) - 2/(13*16)
= 1 - 2/(1*4) - 2/(4*7) - 2/(7*10) - 2/(10*13) - 2/(13*16)
= 1 - (2/3)(1/1 - 1/4) - (2/3)(1/4 - 1/7) - (2/3)(1/7 - 1/10) - (2/3)(1/10 - 1/13) - (2/3)(1/13 - 1/16)
= 1 - (2/3)(1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16)
= 1 - (2/3)(1/1 - 1/16)
= 1 - (2/3)(15/16)
= 1 - 5/8
= 3/8
= 1 - 2/(1*4) - 2/(4*7) - 2/(7*10) - 2/(10*13) - 2/(13*16)
= 1 - (2/3)(1/1 - 1/4) - (2/3)(1/4 - 1/7) - (2/3)(1/7 - 1/10) - (2/3)(1/10 - 1/13) - (2/3)(1/13 - 1/16)
= 1 - (2/3)(1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16)
= 1 - (2/3)(1/1 - 1/16)
= 1 - (2/3)(15/16)
= 1 - 5/8
= 3/8
收錄日期: 2021-04-11 21:30:33
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