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
数学人气:220 ℃时间:2019-08-20 12:32:22
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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=12an=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^nan=3^n-2^n证明...
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