设数列{an}的前几项和为Sn,满足2Sn=an+1-2^n+1+1,且a1,a2+5.a3成等差数列 (1）求a1的值
1个回答
1.
n=1时,
2S1=2a1=a2-2^2 +1
a2=2a1+3
n=2时,
2S2=2(a1+a2)=a3-2^3 +1
a3=2(a1+a2)+7=2(a1+2a1+3)+7=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
2.
n≥2时,
2an=2Sn-2S(n-1)=a(n+1)-2^(n+1)+1-(an-2^n +1)
a(n+1)=3an+2^n
a(n+1)+2^(n+1)=3an+3×2^n=3×(an+2^n)
[a(n+1)+2^(n+1)]/(an +2^n)=3,为定值
a1+2=1+2=3,数列{an +2^n}是以3为首项,3为公比的等比数列
an+2^n=3×3^(n-1)=3^n
an=3^n -2^n
数列{an}的通项公式为an=3^n -2^n
3.
3>2>1,3^n>2^n,an恒>0
[1/a(n+1)]/(1/an)
=an/a(n+1)
=(3^n -2^n)/[3^(n+1)-2^(n+1)]
=(1/3)[3^(n+1)-3×2^n]/[3^(n+1)-2^(n+1)]
=(1/3)[3^(n+1)-2^(n+1)-2^n]/[3^(n+1)-2^(n+1)]
=1/3 - (1/3)×2^n/[3^(n+1)-2^(n+1)]
(1/3)×2^n/[3^(n+1)-2^(n+1)]>0
1/3 -(1/3)×2^n/[3^(n+1)-2^(n+1)]
相关问题
