作业帮谁般偶求下,感激不尽1/(1*2*3)+1/(2*3*4)+1/（3*4*5）...+1/（100*101*102）注意到1/n*(n+1)*(n+2)=1/2*[1/n*(n+1)-1/(n+1)(n+2)]= 1/2{[1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}=1/2*[1/n-2/(n+1)+1/(n+2)] 所以原式=1/2[(1-1/101)-(1/2-1/102)]=50/101-2
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/06 09:54:50
![谁般偶求下,感激不尽1/(1*2*3)+1/(2*3*4)+1/（3*4*5）...+1/（100*101*102）注意到1/n*(n+1)*(n+2)=1/2*[1/n*(n+1)-1/(n+1)(n+2)]= 1/2{[1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}=1/2*[1/n-2/(n+1)+1/(n+2)] 所以原式=1/2[(1-1/101)-(1/2-1/102)]=50/101-2](/uploads/image/z/753202-10-2.jpg?t=%E8%B0%81%E8%88%AC%E5%81%B6%E6%B1%82%E4%B8%8B%2C%E6%84%9F%E6%BF%80%E4%B8%8D%E5%B0%BD1%2F%281%2A2%2A3%29%2B1%2F%282%2A3%2A4%29%2B1%2F%EF%BC%883%2A4%2A5%EF%BC%89...%2B1%2F%EF%BC%88100%2A101%2A102%EF%BC%89%E6%B3%A8%E6%84%8F%E5%88%B01%2Fn%2A%28n%2B1%29%2A%28n%2B2%29%3D1%2F2%2A%5B1%2Fn%2A%28n%2B1%29-1%2F%28n%2B1%29%28n%2B2%29%5D%3D+1%2F2%7B%5B1%2Fn-1%2F%28n%2B1%29%5D-%5B1%2F%28n%2B1%29-1%2F%28n%2B2%29%5D%7D%3D1%2F2%2A%5B1%2Fn-2%2F%28n%2B1%29%2B1%2F%28n%2B2%29%5D+%E6%89%80%E4%BB%A5%E5%8E%9F%E5%BC%8F%3D1%2F2%5B%281-1%2F101%29-%281%2F2-1%2F102%29%5D%3D50%2F101-2)
谁般偶求下,感激不尽1/(1*2*3)+1/(2*3*4)+1/（3*4*5）...+1/（100*101*102）注意到1/n*(n+1)*(n+2)=1/2*[1/n*(n+1)-1/(n+1)(n+2)]= 1/2{[1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}=1/2*[1/n-2/(n+1)+1/(n+2)] 所以原式=1/2[(1-1/101)-(1/2-1/102)]=50/101-2
谁般偶求下,感激不尽1/(1*2*3)+1/(2*3*4)+1/（3*4*5）...+1/（100*101*102）
注意到1/n*(n+1)*(n+2)=1/2*[1/n*(n+1)-1/(n+1)(n+2)]=
1/2{[1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}=1/2*[1/n-2/(n+1)+1/(n+2)]
所以原式=1/2[(1-1/101)-(1/2-1/102)]=50/101-25/102
谁是n呢
谁般偶求下,感激不尽1/(1*2*3)+1/(2*3*4)+1/（3*4*5）...+1/（100*101*102）注意到1/n*(n+1)*(n+2)=1/2*[1/n*(n+1)-1/(n+1)(n+2)]= 1/2{[1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}=1/2*[1/n-2/(n+1)+1/(n+2)] 所以原式=1/2[(1-1/101)-(1/2-1/102)]=50/101-2
注意到1/n*(n+1)*(n+2)=1/2*[1/n*(n+1)-1/(n+1)(n+2)]=
1/2{[1/n-1/(n+1)]-[1/(n+1)-1/(n+2)]}=1/2*[1/n-2/(n+1)+1/(n+2)]
所以原式=1/2[(1-1/101)-(1/2-1/102)]=50/101-25/102
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