S
证明:设等差数列{an}的首项为a1,公差为d, 则Sn=na1+ n(n−1)d 2. bn= Sn n=a1+ n−1 2d. 则bn+1−bn=a1+ n 2d−a1− n−1 2d= d 2. ∴数列{bn}是等差数列.
已知等差数列{an}的前n项和Sn,且bn=S
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