n=1,显然成立,略
假设n=k成立,k≥1
即0*2+1*3+……+(k-1)(k+1)=k(k-1)(2k+5)/6
则n=k+1时
0*2+1*3+……+(k-1)(k+1)+k(k+2)=k(k-1)(2k+5)/6+k(k+2)
=k(2k²+3k-5+6k+12)/6
=k(2k²+9k+7)/6
=k(k+1)(2k+7)/6
=[(k+1)-1](k+1)[2(k+1)+5]/6
所以n属于正整数时,0*2+1*3+……+(n-1)(n+1)=n(n-1)(2n+5)/6
n=1,显然成立,略
假设n=k成立,k≥1
即0*2+1*3+……+(k-1)(k+1)=k(k-1)(2k+5)/6
则n=k+1时
0*2+1*3+……+(k-1)(k+1)+k(k+2)=k(k-1)(2k+5)/6+k(k+2)
=k(2k²+3k-5+6k+12)/6
=k(2k²+9k+7)/6
=k(k+1)(2k+7)/6
=[(k+1)-1](k+1)[2(k+1)+5]/6
所以n属于正整数时,0*2+1*3+……+(n-1)(n+1)=n(n-1)(2n+5)/6