Question

设数列{an}的前n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
（2）求数列{an}的通项公式
（3）证明：对一切正整数n,有1/a1+1/a2+...1/an<3/2

急救！哪位老是教一下 网上查过 看不懂啊~

Answer 1

a1,a2+5,a3成等差数列
a1+a3 = 2(a2+5) (1)

2Sn=a(n+1)-2^(n+1)+1
n=1
2a1 = a2- 4+1
a2= 2a1+3 (2)

n=2
2(a1+a2)=a3-8+1
a3= 2(a1+a2) +7
= 2(a1 +2a1+3) +7
= 6a1+13 (3)
sub (3) ,(2) into (1)
a1 +(6a1+13) = 2(2a1+3+5)
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{2^(n+1)} - 1/2= 3(Sn + 2{2^n} -1/2)
[S(n+1) +2{2^(n+1)} - 1/2]/(Sn + 2{2^n} -1/2) = 3
(Sn + 2{2^n} -1/2)/(S1 + 2{2^1} -1/2) = 3^(n-1)
Sn + 2{2^n} -1/2 = (3/2).3^n
Sn = 1/2 - 2^(n+1) + (3/2).3^n
an = Sn-S(n-1)
= -2^n + 3^n