设等比数列{an}的前n项和为Sn,a4=a1-9,a5,a3,a4成等差数列.(1)求数列{an}的通项公式,(2)证明:对任意k∈N+,Sk+2,Sk,Sk
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设等比数列{an}的前n项和为Sn,a4=a1-9,a5,a3,a4成等差数列. (1)求数列{an}的通项公式, (2)证明:对任意k∈N+,Sk+2,Sk,Sk+1成等差数列. |
答案
(1)设等比数列{an}的公比为q, 则 | a1•q3=a1-9 | 2a1q2=a1q3+a1q4 |
| | ,解得, 故数列{an}的通项公式为:an=(-2)n-1, (2)由(1)可知an=(-2)n-1, 故Sk==, 所以Sk+1=,Sk+2=, ∴Sk+1+Sk+2=+= ==, 而2Sk=2===, 故Sk+1+Sk+2=2Sk,即Sk+2,Sk,Sk+1成等差数列 |
举一反三
若{an}是等差数列,则a1+a2+a3,a4+a5+a6,a7+a8+a9( )A.不是等差数列 | B.是递增数列 | C.是等差数列 | D.是递减数列 |
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