(本小题满分12分)已知数列{an}的前n项和为Sn, 且满足条件:4S n =+ 4n – 1 , nÎN*.(1) 证明:(a n– 2)2 –="0" (
题型:不详难度:来源:
(本小题满分12分)已知数列{an}的前n项和为Sn, 且满足条件:4S n =+ 4n – 1 , nÎN*. (1) 证明:(a n– 2)2 –="0" (n ³ 2);(2) 满足条件的数列不惟一,试至少求出数列{an}的的3个不同的通项公式 . |
答案
(2) 当a1 =1且a n + an – 1 = 2时,得an ="1. " 2)当a1 =1且a n – a n – 1 =" 2" 时,得an =" 2n–1" . 3)当a1 =3且a n – a n – 1 =" 2" 时,得an =" 2n" + 1 . 4)当a1 =3且a n + an – 1 = 2时,得an =2(–1)n+ 1 + 1. |
解析
(1) 由条件4S n =+ 4n – 1 , nÎN*.得4S n – 1 =+ 4(n – 1 ) – 1, 相减得:4a n = – + 4,化成–4a n+ 4–= 0, ∴ (a n– 2)2 –="0" . 4分 (2) 由(1)得:(a n –2 + an – 1 )(a n –2 – a n – 1 ) =" 0∴" a n + an – 1 =" 2 " 或a n – a n – 1 =" 2" . 2分 在4S n =+ 4n – 1中,令n = 1,得4a1 =+ 4 – 1,解得:a1 =1或 a1 ="3. " 2分 分四种情况: 1)当a1 =1且a n + an – 1 = 2时,得an =1. 2)当a1 =1且a n – a n – 1 =" 2" 时,得an =" 2n–1" . 3)当a1 =3且a n – a n – 1 =" 2" 时,得an =" 2n" + 1 . 4)当a1 =3且a n + an – 1 = 2时,得an =2(–1)n+ 1 + 1. 每个1分,有3个即可 |
举一反三
(本小题满分12分)等差数列的前项和为. ⑴求数列的通项与前项和;⑵设,求证:数列中任意不同的三项都不可能成为等比数列. |
(本小题满分14分)已知的首项为a1,公比q为正数(q≠1)的等比数列,其前n项和为Sn,且. (1)求q的值; (2)设,请判断数列能否为等比数列,若能,请求出a1的值,否则请说明理由. |
(本小题满分15分)已知二次函数满足条件:① ; ② 的最小值为. (1) 求函数的解析式; (2) 设数列的前项积为, 且, 求数列的通项公式; (3) 在(2)的条件下, 求数列的前项的和. |
((本小题满分12分) 已知数列,设,数列.(Ⅰ)求数列的通项公式;(Ⅱ)若数列的前项和为,求. |
(本题满分16满分)设正项数列的前项和为,为非零常数.已知对任意正整数,当时,总成立. (1)证明:数列是等比数列;(2) 若正整数成等差数列,求证:≥. |
最新试题
热门考点