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(2)求数列{an}的通项公式
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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(2)求数列{an}的通项公式
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
(2)求数列{an}的通项公式
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(2)求数列{an}的通项公式
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
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
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