麻烦告诉下这题目怎么解
(1-1/2+1/3-1/4......+1/1997-1/1998+1/1999)/[1/(1+1999)+1/(2+2000)+1/(3+2001)......+1/...
(1-1/2+1/3-1/4......+1/1997-1/1998+1/1999)/[1/(1+1999)+1/(2+2000)+1/(3+2001)......+1/(999+2997)+1/(1000+2998)]麻烦高手告诉下计算过程 第一次来这里求助 谢谢了
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(1-1/2+1/3-1/4+…+1/1997-1/1998+1/1999)
/[1/(1+1999)+1/(2+2000)+1/(3+2001)+1/(4+2002)+…+1/(999+2997)+1/(1000+2998)]
=[(1+1/2+1/3+1/4+…+1/1998+1/1999)-(1/2+1/4+1/6+…+1/1996+1/1998)×2]
/[1/2000+1/2002+1/2004+1/2006+…+1/2996+1/2998]
=[(1+1/2+1/3+1/4+…+1/1998+1/1999)-(1+1/2+1/3+…+1/998+1/999)] /[1/(2×1000)+1/(2×1001)+1/(2×1002)+…+1/(2×1998)+1/(2×1999)]
=(1/1000+1/1001+…+1/1998+1/1999) /[(1/1000+1/1001+…+1/1998+1/1999)×1/2]
=2
/[1/(1+1999)+1/(2+2000)+1/(3+2001)+1/(4+2002)+…+1/(999+2997)+1/(1000+2998)]
=[(1+1/2+1/3+1/4+…+1/1998+1/1999)-(1/2+1/4+1/6+…+1/1996+1/1998)×2]
/[1/2000+1/2002+1/2004+1/2006+…+1/2996+1/2998]
=[(1+1/2+1/3+1/4+…+1/1998+1/1999)-(1+1/2+1/3+…+1/998+1/999)] /[1/(2×1000)+1/(2×1001)+1/(2×1002)+…+1/(2×1998)+1/(2×1999)]
=(1/1000+1/1001+…+1/1998+1/1999) /[(1/1000+1/1001+…+1/1998+1/1999)×1/2]
=2
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