1*n+2(n-1)+3(n-2)+······+n*1=1/6*n(n+1)(n+2)

Zhang_HuaShan 1年前 已收到2个回答 举报

太阳买卖他 幼苗

共回答了21个问题采纳率:90.5% 举报

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……

1年前

5

hewl1977 幼苗

共回答了7个问题 举报

一、n=1时,1=1成立
二、令n=k时上式成立,则n=k+1时
1*(n+1)+2*n+3(n-1)+······+n*2+(n+1)*1
=1*n+1+2(n-1)+2+3(n-2)+3+······+n*1+n+(n+1)*1
=1*n+2(n-1)+3(n-2)+······+n*1+1+2+3+······+(n+1)
=1/6*n(n+1)(n+...

1年前

1
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