来自陆元成的问题
【1*n+2(n-1)+3(n-2)+······+n*1=1/6*n(n+1)(n+2)】
1*n+2(n-1)+3(n-2)+······+n*1=1/6*n(n+1)(n+2)
1回答
2020-04-1418:28
【1*n+2(n-1)+3(n-2)+······+n*1=1/6*n(n+1)(n+2)】
1*n+2(n-1)+3(n-2)+······+n*1=1/6*n(n+1)(n+2)
当n=1时显然成立
设当n=k时,有1*k+2(k-1)+3(k-2)+······+k*1=1/6*k(k+1)(k+2)
当n=k+1时,有
1*(k+1)+2(k+1-1)+3(k+1-2)+······+(k+1)*1
=1*k+2(k-1)+3(k-2)+······+k*1+(1+2+3+……+k+(k+1))
=1/6*k(k+1)(k+2)+(k+1)(k+2)/2
=1/6*(k+1)(k+2)(k+3)
即n=k+1时成立
故有……