已知数列｛an｝的前n项和Sn=n^2-9n,则其通项an=多少；若它的第k项满足5＜ak＜8,则k=多少
人气：218 ℃ 时间：2020-01-28 07:35:53
解答
Sk=k^2-9k
S(k-1)=(k-1)^2-9(k-1)=k^2-11k+10
ak=Sk-S(k-1)=k^2-9k-(k^2-11k+10)
=2k-10
所以5
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