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对于任意的正整数n,有1/1*2*3 + 1/2*3*4 +...1/n(n+1)(n+2)

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对于任意的正整数n,有1/1*2*3 + 1/2*3*4 +...1/n(n+1)(n+2)
对于任意的正整数n,有1/1*2*3 + 1/2*3*4 +...1/n(n+1)(n+2)
解 1/n(n+1)(n+2)
=[1/n-1/n+1](1/n+1)
=1/n(n+1)-1/(n+1)(n+2)
=1/2(1/n-1/n+2)-1/n+1+1/n+2
=1/2[1/n+1/n+2-2/n+1]
=1/2[1/n-1/n+1+1/n+2-1/n+1]
所以 1/1*2*3 +1/2*3*4+...+1/n(n+1)(n+2)
=1/2(1-1/2+1/3-1/2+1/2-1/3+1/4-1/3+1/3-1/4+1/5-1/4+...+1/n-1/n+1+1/n+2-1/n+1)
=1/2[1-1/2+1/(n+1)(n+2)]
=1/4-1/2(n+1)(n+2)
因为1/2(n+1)(n+2)>0 所以-1/2(n+1)(n+2)