1/1×2×3+1/2×3×4+…1/n(n+1)(n+2)<1/4
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1/1×2×3+1/2×3×4+…1/n(n+1)(n+2)<1/4
证明这个式子,小于四分之一
证明这个式子,小于四分之一
因为我们学过“裂项法”:1/[n(n+1)]=1/n-1/(n+1),
借用这种“裂项方法”后相互抵消的思想,
可用补充公式:1/[n(n+1)(n+2)]=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}
借用有规律的相邻几项依次抵消,得和为:Sn=…………=1/2{1/2-1/[(n+1)(n+2)] }
备注:
补充公式的来由是 1/[n(n+1)(n+2)]=1/2{(n+2-n)/[n(n+1)(n+2)]}=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}
借用这种“裂项方法”后相互抵消的思想,
可用补充公式:1/[n(n+1)(n+2)]=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}
借用有规律的相邻几项依次抵消,得和为:Sn=…………=1/2{1/2-1/[(n+1)(n+2)] }
备注:
补充公式的来由是 1/[n(n+1)(n+2)]=1/2{(n+2-n)/[n(n+1)(n+2)]}=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}
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