设数列an是公差不为零的等差数列,sn是数列an的前n项和,且s3^2=9s2,s4=4s2,
求数列an的通向公式
人气:166 ℃ 时间:2020-02-03 15:45:05
解答
(a1 + a2 + a3)^2 = 9 (a1 + a2)
(a1 + a2 + a3 + a4) = 4(a1 + a2)
设 公差为d, 则
(a2 - d + a2 + a2 + d)^2 = 9(a2 - d + a2)
(a2 -d + a2 + a2+d + a2 + 2d) = 4(a2 - d + a2)
9a2^2 = 9 (2a2 - d)
4a2 + 2d = 4(2a2 -d)
a2^2 = 2a2 -d
2a2 = 3d
a^2 = 2a2 - 2a2 /3
a^2 = 4a2 /3
a2 = 0 或 4/3
a2 = 0 时, d = 0, 整个数列为0数列, 舍去
a2 = 4/3时, d = 8/9
a1 = a2 - d = 4/9
an = a1 + (n-1)d = 4/9 + 8(n-1)/9 = 4(2n-1)/9
------------------
附录 检验:
a1 = 4/9
a2 = 12/9
a3 = 20/9
a4 = 28/9
S2 = 16/9
S4 = 64/9
S4 = 4S2 成立
S3 = 36/9 = 4
S3^2 = 16
9S2 = 16
S3^2 = 9S2 成立
打了不少字, 楼主别忘记采纳哦.
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