Question

1xn+2x(n-1)+3x(n-2)+.+nx1 求和,怎么算?要方法

Answer 1

1xn+2x(n-1)+3x(n-2)+...+nx1=1\6[n(n+1)(n+2)]
①当n=1时,左边=1=右边显然成立.
②假设当n=k时等式成立,则当n=k+1时,有
左边=1*n+2*（n-1）+3*（n-2）+……+（n-1）*2+n*1
=1*（k+1）+2*k+3*（k-1）+……+k*2+（k+1）*1
=1*k+1+2*（k-1）+2+3*（k-2）+3+……+k*1+k+k+1
=1*k+2*（k-1）+3*（k-2）+……+（k-1）*2+k*1+1+2+3+……+k+(k+1)
=[k(k+1)(k+2)]/6+(k+1)(k+2)/2 注意：这一步左边一半用的是归纳假设,右一半是等差数列
=[(k+1)(k+2)(k+3)]/6
=[n(n+1)(n+2)]/6
=右边成立