Question

数列{an}中,a1=3,且S(n+1)+Sn=2a(n+1),求①数列{an}的通项公式②数列{an}的前n项的和

Answer 1

S(n+1)+Sn=2a(n+1),
sn+s(n-1)=2an(n>=2)

上减下
a(n+1)+an=2a(n+1)-2an
3an=a(n+1)
a(n+1)/an=3

当n=1时，s2+s1=2a2
a1+a2+a1=2a2
2a1=a2
a2/a1=2,a1=3,a2=6
an= 3 n=1
=6*3^n-2 n>=2

sn=3+(6-3^n-2)/1-3
=3+(-3)+3^n-2/2
=3^(n-2)/2

Answer 2

S2+S1=a2+2S1=2 a2, a2=2S1=6, S2=3S1=9
S3+S2=a3+2S2=2a3, a3=2S2=18,S3=3S2=27
a(n+1)=2Sn=2*3^n, n>2
an=2*3^(n-1), n>1, a1=3
Sn=3^n