n(n+1)=1/3[n(n+1)(n+2)-(n-1)n(n+1)]言尽于此.
左边=2(2C2+3C2+……+(n+1)C2) =2(3C3+3C2+4C2……+(n+1)C2) =2(4C3+4C2……+(n+1)C2) =2((n+1)C3+(n+1)C2) =2 (n+2)C3 =n(n+1)(n+2)/3=右边