用数学归纳法证明:1·2·3+2·3·4+3·4·5+.+n(n+1)(n+2)=1/4n(n+1)(n+2)(n+3)

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用数学归纳法证明:1·2·3+2·3·4+3·4·5+.+n(n+1)(n+2)=1/4n(n+1)(n+2)(n+3)
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用数学归纳法证明:1·2·3+2·3·4+3·4·5+.+n(n+1)(n+2)=1/4n(n+1)(n+2)(n+3)
用数学归纳法证明:1·2·3+2·3·4+3·4·5+.+n(n+1)(n+2)=1/4n(n+1)(n+2)(n+3)

用数学归纳法证明:1·2·3+2·3·4+3·4·5+.+n(n+1)(n+2)=1/4n(n+1)(n+2)(n+3)
证:
(1)n=1时,左式=1·2·3=6
右式=1/4·1·2·3·4=6
成立!
(2)假设n=k≥2(k∈N)时成立,即:
1·2·3+2·3·4+3·4·5+.+k(k+1)(k+2)=1/4·k(k+1)(k+2)(k+3)
则当n=k+1时
1·2·3+2·3·4+3·4·5+.+k(k+1)(k+2)+(k+1)(k+2)(k+3)
=(1/4)·k(k+1)(k+2)(k+3)+(k+1)(k+2)(k+3)
=(1/4)·(k+1)(k+2)(k+3)(k+4)
显然成立!
综上,等式对任意n∈N时均成立!