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!