数列an,bn满足bn=a1+2a2+3a3...nan\1+2+3+...n,若bn是等差数列,求证an是等差数列

数列an,bn满足bn=a1+2a2+3a3...nan\1+2+3+...n,若bn是等差数列,求证an是等差数列
数列an,bn满足bn=a1+2a2+3a3...nan\1+2+3+...n,若bn是等差数列,求证an是等差数列

数列an,bn满足bn=a1+2a2+3a3...nan\1+2+3+...n,若bn是等差数列,求证an是等差数列
证明：
先对式子进行化简：a1+2a2+3a3...+nan=bn*(1+2+3+...+n)=bn*n(n+1)/2
取n-1项,故有a1+2a2+3a3...+(n-1)a(n-1)=b(n-1)*n(n-1)/2
两个式子对应左右相减得到：nan=bn*n(n+1)/2-b(n-1)*n(n-1)/2
两边除以n,得an=bn*(n+1)/2-b(n-1)*(n-1)/2=[(n+1)bn-(n-1)b(n-1)]/2
由假设,bn是等差数列,不妨设bn-b(n-1)=d（常数）,
故an=[nd+bn+b(n-1)]/2
从而an-a(n-1)=3d/2,即an为等差数列.