1*N+2*(N-1)+3*(N-2)+...+N*1=1/6N(N+1)(N+2)
问题描述:
1*N+2*(N-1)+3*(N-2)+...+N*1=1/6N(N+1)(N+2)
答
第N项
1*N+2*(N-1)+3*(N-2)+...+N*1=1/6N(N+1)(N+2)
第N+1项
1*(N+1)+2*(N+1-1)+3*(N+1-2)+...+(N+1)*1
=1*N+1+2*(N-1)+2+3*(N-2)+3+...+N*1+1
=1*N+2*(N-1)+3*(N-2)+...+N*1+(1+2+3+...+n/2+n/2-1+...+1)
=1/6N(N+1)(N+2)+(n+1)*(n+2)/2
=1/6(N+1)(N+2)(N+3)
1+2+3+...+n-1+n=n*(n+1)/2
答
楼上再N+1证时有错误,正确的如下:用数学归纳法:当n=1时左边=1=右边假设n=N时成立,则n=N+1时:1*(N+1)+2*(N+1-1)+3*(N+1-2)+...+N*(1+1)+(N+1)*1=1*N+1 + 2*(N-1)+2 + 3*(N-2)+3+...+N*1+N + (N+1)=1*N+2*(N-1)+3...