证明:1*n+2*(n-1)+3*(n-2)+…+n*1=n*(n+1)*(n+2)/61`当n=1时左边=1*1=1右边=1*(1+1)*(1+2)/6=1等式成立.2`假设当n=k时等式成立,即1*k+2*(k-1)+3*(k-2)+…+k*1=1*k+2*(k-1)+3*(k-2)+…+(k-1)[k-(k-2)]+k*[k-(k-1)]=k*(k+1)*(k+2)/6当n=k+1时左边=1 展开
证明:1*n+2*(n-1)+3*(n-2)+…+n*1=n*(n+1)*(n+2)/61`当n=1时左边=1*1=1右边=1*(1+1)*(1+2)/6=1等式成立.2`假设当n=k时等式成立,即1*k+2*(k-1)+3*(k-2)+…+k*1=1*k+2*(k-1)+3*(k-2)+…+(k-1)[k-(k-2)]+k*[k-(k-1)]=k*(k+1)*(k+2)/6当n=k+1时左边=1*(k+1)+2*(k-1+1)+3*(k-2+1)+…+k[k+1-(k-2+1)]+(k+1)*[k+1-(k-1+1)]=1*(k+1)+2*(k-1+1)+3*(k-2+1)+…+k*(1+1)+(k+1)*1=1*(k+1)+2*(k-1+1)+3*(k-2+1)+…+k*1+k+k+1=[1*k+2*(k-1)+3*(k-2)+…+k*1]+[1+2+3+…+k+(k+1)]=k*(k+1)*(k+2)/6+(k+1)*(k+2)/2=(k+1)*(k+2)*(k+3)/6右边=(k+1)*(k+2)*(k+3)/6所以左边=右边综上所述:1*n+2*(n-1)+3*(n-2)+…+n*1=n*(n+1)*(n+2)/6,即命题得证 收起
