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
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![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](/uploads/image/z/3618698-50-8.jpg?t=%E6%95%B0%E5%88%97an%2Cbn%E6%BB%A1%E8%B6%B3bn%3Da1%2B2a2%2B3a3...nan%5C1%2B2%2B3%2B...n%2C%E8%8B%A5bn%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%B1%82%E8%AF%81an%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97)
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.