设{an}是等差数列,求证以b=(a1+a2+a3+...+an)/n为通项公式的数列{bn}是等差数列
问题描述:
设{an}是等差数列,求证以b=(a1+a2+a3+...+an)/n为通项公式的数列{bn}是等差数列
答
a1+a2+a3+…+an=na1+[n(n-1)d]/2,则bn=a1+(d/2)(n-1),从而
b(n+1)-bn=[a1+(d/2)n]-[a1+(d/2)(n-1)]=d/2=常数,则数列{bn}是等差数列.