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

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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

你好很高兴回答你的问题
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

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