解ï¼
n(n+1)(n+2)(n+3)=(1/5)Ã[n(n+1)(n+2)(n+3)(n+4)-(n-1)n(n+1)(n+2)(n+3)]
1Ã2Ã3Ã4+2Ã3Ã4Ã5+...+27Ã28Ã29Ã30
=(1/5)Ã(1Ã2Ã3Ã4Ã5-0Ã1Ã2Ã3Ã4+2Ã3Ã4Ã5Ã6-1Ã2Ã3Ã4Ã5+...+27Ã28Ã29Ã30Ã31-26Ã27Ã28Ã29Ã30)
=(1/5)Ã26Ã27Ã28Ã29Ã30
=3420144
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