简单,首先声明一下n^2(n+1)^2是整数,[x]是整数,所以[[n^2(n+1)^2]]=n^2(n+1)^2,[[x]]=[x]!所以等式两边同时取整,则[x+2[x]+3[x]+4[x]+…+n[x]]=[x]+2[[x]]+3[[x]]+...n[[x]]=[x]+2[x]+3[x]+4[x]+…+n[x]=[n²(n+1)²/2
]=n^2(n+1)^2=x+2[x]+3[x]+4[x]+…+n[x],即[x]+2[x]+3[x]+4[x]+…+n[x]=x+2[x]+3[x]+4[x]+…+n[x],消去之后就是x=[x]!