证明:
你的题目缺少一个条件,即{an}是等差数列,公差为d,此时:
∵数列{an}是等差数列
∴an-a(n-1)=d
则 wan+b-w[a(n-1)+b]
=wan-wa(n-1)
=w[an-a(n-1)]
=wd
∴ 数列{w an+b}(w,b是常数)是公差为wd的等差数列.
证明:
你的题目缺少一个条件,即{an}是等差数列,公差为d,此时:
∵数列{an}是等差数列
∴an-a(n-1)=d
则 wan+b-w[a(n-1)+b]
=wan-wa(n-1)
=w[an-a(n-1)]
=wd
∴ 数列{w an+b}(w,b是常数)是公差为wd的等差数列.