Question

数列an中,a1=3,a2=5,其前n项和为Sn,满足Sn+S(n-2)=2S(n-1)+2^（n-1),n>=3,求a

Answer 1

n≥3时,

Sn+S(n-2)=2S(n-1)+2^(n-1)

Sn-S(n-1)-2ⁿ=S(n-1)-S(n-2)-2^(n-1)

an -2ⁿ=a(n-1)-2^(n-1)

a1-2=3-2=1

数列{an-2ⁿ}是各项均为1的常数数列.

an-2ⁿ=1

an=2ⁿ+1

n=1时,a1=2+1=3；n=2时,a2=2²+1=4+1=5,同样满足.

数列{an}的通项公式为an=2ⁿ+1.