题抄错了吧，是不是1*2+2*3+3*4……n（n+1）=？
数列求和,把每一项n*(n+1)分成n*n和n
那么1*2+2*3+3*4……+n(n+1)
=(1*1+2*2+3*3+……+n*n)+(1+2+3+……+n)
=(1/6)*n*(n+1)(2n+1)+(1/2)*n(n+1)
=(1/6)*n*(n+1)(2n+4)
=(1/3)n(n+1)(n+2)
1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2)
=1/4[1*2*3*4]+1/4[2*3*4*5-1*2*3*4]+1/4[3*4*5*6-2*3*4*5]+...+1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
=1/4[1*2*3*4+2*3*4*5-1*2*3*4+3*4*5*6-2*3*4*5+...+n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
=1/4*n(n+1)(n+2)(n+3)
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ号：24596024