Question

设数列{an}的前n项和为Sn,已知a1=a,a(n+1)=Sn+3^n,n∈N

(1)设bn=Sn 3^n,求数列{bn}的通项公式;
(2)若a(n+1)≥an,n∈N,求a的取值范围.

Answer 1

a(2)=a1+3^1=a+3
S(n+1)=Sn+a(n+1)=2Sn+3^n
S(n+1)-3^(n+1)=2(Sn-3^n)
b(n+1)=2bn
bn成等比数列
b1=a1-3=a-3
bn=0.5*(a-3)*2^n
Sn=bn+3^n=0.5*(a-3)*2^n+3^n
a(n+1)=0.5*(a-3)*2^n+2*3^n
an=0.25*(a-3)*2^n+2/3*3^n
a(n+1)>=an
0.5*(a-3)*2^n+2*3^n>=0.25*(a-3)*2^n+2/3*3^n
0.25(a-3)*2^n+4/3*3^n>=0
a-3>=-16/3*1.5^n
a>=3-16/3*1.5^n

Answer 2

a(n+1)=Sn+3^n--------------------1
an=s(n-1)+3^n-1----------------2
1-2, an+1=2an-2*3^n-1
变形得到a(n+1)+2*3^n=2（an+2*3^n-1）
所以an+2*3^n-1是等比数列
a1+2=a+2
an+2*3^n-1=（a+2）*2^n-1
所以an=（a+2）*2^n-1-2*3^n-1