计算,化简SOS计算1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004)(a+2005)

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计算,化简SOS计算1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004)(a+2005)
计算,化简SOS
计算1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004)(a+2005)

计算,化简SOS计算1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004)(a+2005)
1/a(a+1)=1/a-1/(1+a)
1/(a+1)(a+2)=1/(1+a)-1/(2+a)
1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+.+1/(a+2004)(a+2005)
=1/a-1/(a+2005)

1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+......+1/(a+2004)(a+2005)
=[1/a-1/(a+1)]+[1/(a+1)-1/(a+2)]+.....+[1/(a+2004)-1/(a+2005) ]
=1/a-1/(a+2005)
=2005/[a(a+2005)]

1/(a+n)(a+n+1)=1/(a+n)-1/(a+n+1)这是一个通项公式,根据这个就可以把上面的式子做出来,结果是1/a-1/(a+2005)=2005/a(a+2005)

2005/a(a+2005]

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