Question

设数列{an}的前n项和为Sn，已知a1=1，Sn+1=4an+2， (Ⅰ)设bn=an/2n，证明数列{bn}是等产数列；

（Ⅱ）求数列{an}的通项公式及前n项和的公式

Answer 1

(I)
S(n+1)=4an+2
n=1
a1+a2=4a1+2
a2=3a1+2=5

a(n+1)=S(n+1)-Sn
=4an -4a(n-1)
a(n+1)-2an = 2[an - 2a(n-1)]
=>{an - 2a(n-1)} 是等比数列, q=2
an - 2a(n-1) = 2^(n-2). (a2 - 2a1)
= 3. 2^(n-2)
an/2^n - a(n-1)/2^(n-1) = 3/4
=>{an/2^n}是等差数列, d=3/4
an/2^n -a1/2 = 3(n-1)/4
an/2^n = (3n-1)/4
an = (3n-1).2^(n-2)

(II)
Sn=4a(n-1)+2
=4(3n-4).2^(n-3) +2
=2 +(3n-4).2^(n-1)