3．数列9,99,999.-的前n项和为( ) A.(10n-1)+n B．10n-1 C.(10n-1) D.(10n-1)-n 解析:选D.∵数列通项an＝10n-1. ∴S——青夏教育精英家教网—

摘要：3．数列9,99,999.-的前n项和为( ) A.(10n-1)+n B．10n-1 C.(10n-1) D.(10n-1)-n 解析:选D.∵数列通项an＝10n-1. ∴Sn＝(10+102+103+-+10n)-n ＝-n ＝(10n-1)-n.故应选D.

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A.(10ⁿ-1)+n B.10ⁿ-1

C.(10ⁿ-1) D.(10ⁿ-1)-n

A. B.10ⁿ-1 C. D.

A.10ⁿ－1 B.(10ⁿ－1)－n

C.(10ⁿ－1) D.(10ⁿ－1)+n

(1)-1,1,-2,2,-3,3,…；(2)9,99,999,9 999,….

(1)-3,0,3,6,9;

(2)3,5,9,17,33;

(3)4,-4,4,-4,4;

(4)1,0,1,0,1;

(5),,-,;

(6)9,99,999,9 999.

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