（1）设数列｛an｝是公差部位零的等差数列,Sn是数列｛an｝前n项的和,用(S3)^2=9*(S2),(S4)=4*(

1.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
Sn=a1*n+n(n-1)d/2
由
S50=200
S100-S50=2700
得出
50*a1+1225*d=200
100*a1+4950*d=2700+200
解得
a1=-20.5
d=1

？？？
不懂

1.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(2a...