设数列AN是公差不=0的等差数列SN是数列AN的前N项的和且S3的平方=9S2,S4=4S2求数列AN的通项式

设该等差数列是首项为a1,公差为d
S3=3a1+3(3-1)*d/2=3a1+3d
S2=2a1+2(2-1)*d/2=2a1+d
S4=4a1+4(4-1)*d/2=4a1+6d
又:S3²=9S2
S4=4S2
所以:(3a1+3d)²=9(2a1+d)........(1)
4a1+6d=4(2a1+d).......(2)
(1)化简得:9a1+18a1*d+9d²=18a1+9d
9d²+18a1*d=a1+d......(3)
(2)化简得:4a1+6d=8a1+4d
d=2a1.......(4)
将(4)代入(3)得:
9*4a1²+18a1*2a1=a1+2a1
72a1²-3a1=0
a1(24a1-1)=0
a1=1/24(a1=0不符合条件,如果a1=0,d=2a1=0,公差为0,与已知相矛盾)
d=2a1=1/12
所以:
通项公式an=1/24+(n-1)/12