题目

设 数列｛a n｝的前n项和为Sn,a1=1 S(n+1)=4an+2(n属于N*)
(1)bn=a(n+1)-2an,求bn
(2)若cn=1/bn,求｛cn｝的前6项和T6
(3)若dn=an/(2^n),证明｛dn｝是等差数列
（a,S,b,c,d,T后或其括号内皆为下标）

提问时间：2021-04-01

答案

1.
S(n+1)=4an+2
S2=4a1+2=6
a2=6-1=5
S(n+1)=4an+2
Sn=4a(n-1)+2
a(n+1)=S(n+1)-Sn=4an-4a(n-1)
a(n+1)-2an=2an-4a(n-1)=2[an-2a(n-1)]
bn=2b(n-1)
bn=b1*2^(n-1)
=(a2-2a1)2^(n-1)
=3*2^(n-1)
2.
cn=1/bn=(1/3)(1/2)^(n-1)
Tn=(1/3)[1-(1/2)^n]/(1-1/2)
=(2/3)[1-(1/2)^n]
T6=(2/3)[1-(1/2)^6]
=(2/3)[1-1/64]
=21/32
3.
dn=an/(2^n)
an=(2^n)dn
bn=a(n+1)-2an=3*2^(n-1)
[2^(n+1)]d(n+1)-2(2^n)dn=3*2^(n-1)
d(n+1)-dn=3*2^(n-1)/2^(n+1)
=3/4
则bn是首项为b1=a1/2=1/2公差为3/4的等差数列.证毕.