Question

已知公差不为零的等差数列{an}的前n项和Sn

Answer 1

设公差为d,an=a1+(n-1)d,Sn=na1+n(n-1)d/2.
已知3Sn=(an)^2+p(an)+q
(1)
所以有3Sn+1=(an+1)^2+p(an+1)+q
(2)
3Sn=1=(an-1)^2+p(an-1)+q
(3)
(2)-(1)得：3a1+3nd=[2a1+(2n-1)d]d+pd
(4)
(1)-(3)得：3a1+3(n-1)d=[2a1+(2n-3)d]
d+pd
(5)
(4)-(5)得：3d=2d^2,因为d不是0，所以d=1.5
代入Sn和an有：an=1.5n-2.5,Sn=3n^2/4-7n/4
再代入3Sn=(an)^2+p(an)+q可得
p=17/2,q=15

Answer 2

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
望采纳