用数学归纳法证明:1/1*2*3+1/2*3*4+...+1/N(N+1)(N+2)=N(N+3)/4(N+1)(N+2)

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用数学归纳法证明:1/1*2*3+1/2*3*4+...+1/N(N+1)(N+2)=N(N+3)/4(N+1)(N+2)
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用数学归纳法证明:1/1*2*3+1/2*3*4+...+1/N(N+1)(N+2)=N(N+3)/4(N+1)(N+2)
用数学归纳法证明:1/1*2*3+1/2*3*4+...+1/N(N+1)(N+2)=N(N+3)/4(N+1)(N+2)

用数学归纳法证明:1/1*2*3+1/2*3*4+...+1/N(N+1)(N+2)=N(N+3)/4(N+1)(N+2)
过程较繁琐,但是道理很清晰
1/n(n+1)(n+2)=1/2n(n+1)-1/2(n+1)(n+2)=1/2(1/n+1/(n+2)-2/(n+1))
利用数学归纳法
先证n=1成立
设n=k成立,证明n=k+1成立

N=1时,左边=1/6=右边
设N=n时等式成立,则
N=n+1时,左边=1/1*2*3+1/2*3*4+...+1/n(n+1)(n+2)+1/(n+1)(n+2)(n+3)
=n(n+3)/4(n+1)(n+2)+1/(n+1)(n+2)(n+3)
=[n(n+3)(n+3)+4]/[4(n+1)(n+2)(n+3)]
=[(n+1)(n+1)(n+4)...

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N=1时,左边=1/6=右边
设N=n时等式成立,则
N=n+1时,左边=1/1*2*3+1/2*3*4+...+1/n(n+1)(n+2)+1/(n+1)(n+2)(n+3)
=n(n+3)/4(n+1)(n+2)+1/(n+1)(n+2)(n+3)
=[n(n+3)(n+3)+4]/[4(n+1)(n+2)(n+3)]
=[(n+1)(n+1)(n+4)]/[4(n+1)(n+2)(n+3)]
=[(n+1)(n+4)]/[4(n+2)(n+3)]=右边
故由数学归纳法,原等式成立

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证明:当n=1式,原式显然成立
(2)假设n=k,
1/1*2*3+1/2*3*4+...+1/k(k+1)(k+2)=k(k+3)/4(k+1)(k+2)
当n=k+1,
1/1*2*3+1/2*3*4+...+1/k(k+1)(k+2)+1/(k+1)(k+2)(k+3)
=k(k+3)/4(k+1)(k+2)+1/(k+1)(k+2)(k+3)

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证明:当n=1式,原式显然成立
(2)假设n=k,
1/1*2*3+1/2*3*4+...+1/k(k+1)(k+2)=k(k+3)/4(k+1)(k+2)
当n=k+1,
1/1*2*3+1/2*3*4+...+1/k(k+1)(k+2)+1/(k+1)(k+2)(k+3)
=k(k+3)/4(k+1)(k+2)+1/(k+1)(k+2)(k+3)
=[k(k+3)^2+4]/4(k+1)(k+2)(k+3)
=(k+4)*(k+1)^2/4(k+1)(k+2)(k+3)
=(k+4)(k+1)/4(k+2)(k+3)
原始成立
综合(1)(2)的证

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我不用数学归纳法,不知道是否对你有帮助,
由1/1×2×3=﹙2/1×2×3﹚×1/2
=1/2﹙1/1×2-1/2×3﹚
1/2×3×4=1/2﹙1/2×3-1/3×4﹚
。。。。。。
1/n﹙n+1﹚﹙n+2﹚=1/2[1/n﹙n+1﹚-1/﹙n+1﹚﹙n+2﹚]
∴原式=1/2[1/1×2-1/﹙n+1﹚﹙n+2﹚]
=n﹙n+3﹚/4...

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我不用数学归纳法,不知道是否对你有帮助,
由1/1×2×3=﹙2/1×2×3﹚×1/2
=1/2﹙1/1×2-1/2×3﹚
1/2×3×4=1/2﹙1/2×3-1/3×4﹚
。。。。。。
1/n﹙n+1﹚﹙n+2﹚=1/2[1/n﹙n+1﹚-1/﹙n+1﹚﹙n+2﹚]
∴原式=1/2[1/1×2-1/﹙n+1﹚﹙n+2﹚]
=n﹙n+3﹚/4﹙n+1﹚﹙n+2﹚

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