已知公差为d的等差数列的前n项之和为Sn,则Sn+1/n+1-Sn/n
人气:342 ℃ 时间:2020-09-02 10:18:01
解答
Sn+1/n+1=[(n+1)a1+n*(n+1)d/2]/(n+1)=a1+n*d/2
Sn/n=[n*a1+n*(n-1)d/2]/n=a1+n*d/2-d/2
Sn+1/n+1-Sn/n =(a1+n*d/2)-(a1+n*d/2-d/2)
=d/2
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