Question

设数列{an}的前n项和为Sn,满足2Sn=an+1 -2^(n+1) + 1 ,n属于N*,且a1、a2 +5、a3

Answer 1

n=1时,2S1=2a1=a2-2²+1=a2-3

a2=2a1+3

n=2时,2S2=2(a1+a2)=2(a1+2a1+3)=6a1+6=a3-2³+1=a3-7

a3=6a1+13

a1、a2+5、a3成等差数列,则

2(a2+5)=a1+a3

2(2a1+3+5)=a1+6a1+13

整理,得

3a1=3

a1=1

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

S(n+1)=3Sn +2^(n+1) -1

S(n+1)+2^(n+2) -1/2=3Sn+3×2^(n+1) -3/2=3[Sn+2^(n+1) -1/2]

[S(n+1)+2^(n+2) -1/2]/[Sn +2^(n+1) -1/2]=3,为定值

S1+2² -1/2=a1+4-1/2=1+4-1/2=9/2,数列{Sn +2^(n+1) -1/2}是以9/2为首项,3为公比的等比数列

Sn +2^(n+1) -1/2=(9/2)×3^(n-1)=(1/2)×3^(n+1)

Sn=(1/2)×3^(n+1) -2^(n+1) +1/2

n≥2时,an=Sn-S(n-1)=(1/2)×3^(n+1) -2^(n+1) +1/2 -[(1/2)×3ⁿ-2ⁿ+1/2]=3ⁿ-2ⁿ

n=1时,a1=3-2=1,同样满足通项公式

数列{an}的通项公式为an=3ⁿ-2ⁿ