两道高中数列题，求详细解答

1

a2=(1/3)S1=(1/3)a1=1/3
a3=(1/3)S2=(1/3)(a1+a2)=4/9
a4=(1/3)S3=(1/3)(a1+a2+a3)=16/27

a(n+1)=(1/3)Sn
a(n+2)=(1/3)S(n+1)
a(n+2)-a(n+1)=(1/3)[S(n+1)-Sn]=(1/3)a(n+1)
a(n+2)/a(n+1)=4/3
所以：{an}是公比为4/3的等比数列
an=a1*q^(n-1)=(4/3)^(n-1)

{a2n}是公比为(4/3)^2=16/9的等比数列，首项为a2=4/9
a2+a4+a6+……+a2n
=(4/9)[(16/9)^n-1}/[(16/9)-1]
=(4/7)[(16/9)^n-1]

2
(1)
已知等差数列 a15=33，a45=153，求a61
a45=a15+30d
153=33+30d
30d=120
d=4

a15=a1+14d
33=a1+14*4
33=a1+56
a1=-23

a61=a1+60d
=-23+60*4
=-23+240
=217
(2)

S8=a1+a2+...a8
=a1+(a1+d)+...(a1+7d)
=8a1+28d=48

=>2a1+7d=12

同样S12=a1+...+a12
=a1+...+(a1+11d)
=12a1+66d=168
=>2a1+11d=28

=>a1=-8 d=4

(3)令等差数列{an},首项为a1，公差为d
则a1=a1 a2=a1+d a3=a1+2d ………… a6=a1+5d=10
S5=a1+a2+a3+a4+a5=a1+a1+d+a1+2d+a1+3d+a1+4d=5*a1+10d=5
得到方程
a1+5d=10
5a1+10d=5
解得a1=-5 d=3
则an=-8+3n --> a8=-8+24=16
根据Sn=(a1+an)*n/2得
S8=(-5+16)*8/2=44

所以a8=16
S8=44

(4)S31=31（a1+a31）/2,
而a1=a16-15d,
a31=a16+15d,

所以原式=31·a16=93