试证:对任意正整数n,有1/(1*2*3)+1/(2*3*4)+…+1/(n(n+1)(n+2))
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试证:对任意正整数n,有1/(1*2*3)+1/(2*3*4)+…+1/(n(n+1)(n+2))
![试证:对任意正整数n,有1/(1*2*3)+1/(2*3*4)+…+1/(n(n+1)(n+2))](/uploads/image/z/18119297-65-7.jpg?t=%E8%AF%95%E8%AF%81%EF%BC%9A%E5%AF%B9%E4%BB%BB%E6%84%8F%E6%AD%A3%E6%95%B4%E6%95%B0n%2C%E6%9C%891%2F%EF%BC%881%2A2%2A3%EF%BC%89%2B1%2F%EF%BC%882%2A3%2A4%EF%BC%89%2B%E2%80%A6%2B1%2F%EF%BC%88n%EF%BC%88n%2B1%EF%BC%89%EF%BC%88n%2B2%EF%BC%89%EF%BC%89)
因为1/(1*2*3)=(1/2)*[1/(1*2)-1/(2*3)],
1/(2*3*4)=(1/2)*[1/(2*3)-1/(3*4)],
...
(可以把右边通分,证明等式成立)
所以1/(1*2*3)+1/(2*3*4)+...+1/n(n+1)(n+2)
=(1/2)*[1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+...+1/n(n+1)-1/(n+1)(n+2)]
=(1/2)*[1/2-1/(n+1)(n+2)]
=1/4-1/2(n+1)(n+2)
因为1/2(n+1)(n+2)>0,所以式子左边=1/4-1/2(n+1)(n+2)
1/(2*3*4)=(1/2)*[1/(2*3)-1/(3*4)],
...
(可以把右边通分,证明等式成立)
所以1/(1*2*3)+1/(2*3*4)+...+1/n(n+1)(n+2)
=(1/2)*[1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+...+1/n(n+1)-1/(n+1)(n+2)]
=(1/2)*[1/2-1/(n+1)(n+2)]
=1/4-1/2(n+1)(n+2)
因为1/2(n+1)(n+2)>0,所以式子左边=1/4-1/2(n+1)(n+2)
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