数列{an}的前n项和Sn,满足2Sn=an+1-2的n+1(n∈N),且a1,a2+5,a3成等差数列
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数列{an}的前n项和Sn,满足2Sn=an+1-2的n+1(n∈N),且a1,a2+5,a3成等差数列
1.求a1的值
2.求数列{an}的通项
3.证明:对一切正整数n,有1/a1+1/a2+…+1/an
1.求a1的值
2.求数列{an}的通项
3.证明:对一切正整数n,有1/a1+1/a2+…+1/an
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
再问: 亲 2^(n+1)+1这里是不是多了一个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
再问: 亲 2^(n+1)+1这里是不是多了一个1?
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