设数列{An}是公差不为零的等差数列,Sn是数列的前n项和,且S3^2=9S2,S4=4S2,则数列{An}的通项公式是

(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

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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

设公差为d，
则S2=a1+a2=2a1+d
S3=a1+a2+a3=3a1+3d
S4=a1+a2+a3+a4=4a1+6d
S3^2=9S2,则(3a1+3d)²=9(2a1+d)，即(a1+d)²=2a1+d
S4=4S2,则4a1+6d=8a1+4d,则2a1=d代入上式，得
(a1+2a1)²=2a1+2a1

由第二式两边减S2得:
a3+a4=3(a1+a2)，
即2a1+5d=6a1+3d，
即d=2a1。
代入第一式得:
(a1+3a1+5a1)^2=9(a1+3a1)
即81a1^2=36a1
得a1=0(舍)，a1=4/9，d=8/9
所以An=8/9n-4/9
我用手机打的，望采纳

设首项为A1公差为d
S2=2A1+d；S3=3A1+3d；S4=4A1+6d
即(3A1+3d)^2=9(2A1+d)，4A1+6d=4(2A1+d)
解得，A1=4/9，d=8/9
An=A1+(n-1)d=4/9+8(n-1)/9=(8n-4)/9

d=a4-a3=(s4-s3)-(s3-s2)=s4+s2=5s2=5(a1+a2)=10a1+5d
(s3)^2=(3a1+3d)^2=9*(2a1+d)
(a1+d)^2=2a1+d
a1^2+d^2+2a1d-2a1-d=0
a1=4/9, d=8/9
然后 通项公式就不用写了吧~~~^_^ 不懂再问呀~~~d=a4-a3=(s4-s3)-(s3-s...

s4=4a1+6d s3=3a1+3d s2=2a1+d
因为s4=4s2所以4a1+6d =4（2a1+d） 得出d=2a1
因为S3^2=9S2所以（3a1+3d)^2= 9(2a1+d) (3a1+3*2a1)^2=9(2a1+2a1)整理81a1^2=36a1
a1=09（舍） a1=4/9 所以d=8/9 An=4/9+(n-1)8/9