作业帮设数列{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,求数列{an}的通项公式；证明：对一切正整数n,有1/a1+1/a2+...1/an
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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,求数列{an}的通项公式；证明：对一切正整数n,有1/a1+1/a2+...1/an
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,求数列{an}的通项公式；证明：对一切正整数n,有1/a1+1/a2+...1/an
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,求数列{an}的通项公式；证明：对一切正整数n,有1/a1+1/a2+...1/an
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2Sn=a（n+1）-2^（n+1）+1令n=1,2联立（a2+5）*2=a1+a3得a1=1
2an=2sn-2sn-1=a（n+1）-an-2^n
即a（n+1)=3an+2^n
所以a（n+1)+2^（n+1）=3*(an+2^n)
an+2^n=(a1+2^1）*3^(n-1)=3^n
an=3^n-2^n
证明只要证1/a1+1/a2+...1/an
a1,a2+5,a3成等差数列
a1+a3 = 2(a2+5)
2a2-a3= a1-10 (1)
2Sn=a(n+1)-2^(n+1)+1
n=2
2S2 = a3- 7
2(a1+a2) = a3-7
2a2-a3 = -2a1-7 ...
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a1,a2+5,a3成等差数列
a1+a3 = 2(a2+5)
2a2-a3= a1-10 (1)
2Sn=a(n+1)-2^(n+1)+1
n=2
2S2 = a3- 7
2(a1+a2) = a3-7
2a2-a3 = -2a1-7
a1-10=-2a1-7 (2a2-a3= a1-10)
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)
{Sn + 2(2^n) -1/2} 是等比数列, q=3
Sn + 2(2^n) -1/2 = 3^(n-1) .(S1 + 4 -1/2)
= (3/2).3^n
Sn = (3/2).3^n - 2(2^n) +1/2
an = Sn - S(n-1)
= 3^n - 2^n
(2)
1/an = 1/(3^n - 2^n)
= 1/[3^(n-1) + 2.3^(n-2)+...+2^(n-1)]
< 1/3^(n-1) ( for n>= 2 )
a^n-b^n = (a-b)[a^(n-1)+a^(n-2).b +....+b^(n-1) ]
1/a1+1/a2+...+1/an
=1+ 1/a2+...+1/an
< 1+ [1/3^1+1/3^2+....+1/3^(n-1) ]
= 1+ (1/2)( 1- 1/3^n)
< 1+1/2
=3/2
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