1/(1*4)=(1/3)*[1-(1/4)]1/(4*7)=(1/3)*[(1/4)-(1/7)]1/(7*10)=(1/3)*[(1/7)-(1/10)].1/[n*(n+3)]=(1/3)[(1/n)-1/(n+3)]则计算:[1/(1*4)]+[1/(4*7)]+[1/(7*10)]+.+[1/(2005*2008)]的值
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1/(1*4)=(1/3)*[1-(1/4)]1/(4*7)=(1/3)*[(1/4)-(1/7)]1/(7*10)=(1/3)*[(1/7)-(1/10)].1/[n*(n+3)]=(1/3)[(1/n)-1/(n+3)]则计算:[1/(1*4)]+[1/(4*7)]+[1/(7*10)]+.+[1/(2005*2008)]的值
1/(1*4)=(1/3)*[1-(1/4)]
1/(4*7)=(1/3)*[(1/4)-(1/7)]
1/(7*10)=(1/3)*[(1/7)-(1/10)]
.
1/[n*(n+3)]=(1/3)[(1/n)-1/(n+3)]
则计算:
[1/(1*4)]+[1/(4*7)]+[1/(7*10)]+.+[1/(2005*2008)]的值
1/(1*4)=(1/3)*[1-(1/4)]1/(4*7)=(1/3)*[(1/4)-(1/7)]1/(7*10)=(1/3)*[(1/7)-(1/10)].1/[n*(n+3)]=(1/3)[(1/n)-1/(n+3)]则计算:[1/(1*4)]+[1/(4*7)]+[1/(7*10)]+.+[1/(2005*2008)]的值
把1/(1*4)、1/(4*7)、1/(7*10)...直到1/(2005*2008)带入式子中得到下式:
(1/3)*[1-(1/4)]+(1/3)*[(1/4)-(1/7)]+(1/3)*[(1/7)-(1/10)]+...+(1/3)*[(1/2005)-(1/2008)]
提取公因式1/3
解得1/3*(1-1/2008)
所以答案为:669/2008
669/2008
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