已知等差数列{an}的首项a1为a(a∈R,a≠0).设数列的前n项和为Sn,且对任意正整数n都有
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已知等差数列{an}的首项a1为a(a∈R,a≠0).设数列的前n项和为Sn,且对任意正整数n都有
且对任意正整数都有a2n/an=(4n-1)/(2n-1).
(1)求数列{an}的通项公式及Sn
(2)是否存在正整数n和k,使得Sn,Sn+1,Sn+k成等比数列?若存在,求出n和k的值,若不存在,请说明理由.
且对任意正整数都有a2n/an=(4n-1)/(2n-1).
(1)求数列{an}的通项公式及Sn
(2)是否存在正整数n和k,使得Sn,Sn+1,Sn+k成等比数列?若存在,求出n和k的值,若不存在,请说明理由.
![已知等差数列{an}的首项a1为a(a∈R,a≠0).设数列的前n项和为Sn,且对任意正整数n都有](/uploads/image/z/6543833-41-3.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%A6%96%E9%A1%B9a1%E4%B8%BAa%28a%E2%88%88R%2Ca%E2%89%A00%29.%E8%AE%BE%E6%95%B0%E5%88%97%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8F%E6%AD%A3%E6%95%B4%E6%95%B0n%E9%83%BD%E6%9C%89)
a(2n)=a1+(2n-1)d
an=a1+(n-1)d
即:(a1+(2n-1)d)/(a1+(n-1)d)=(4n-1)/(2n-1)
即:(2n-1)(a1+(2n-1)d)=(4n-1)(a1+(n-1)d)
即:2na1=nd,即:d=2a1=2a
1
{an}通项:an=a1+(n-1)d=a+2(n-1)a=(2n-1)a
Sn=na1+n(n-1)d/2=na+n(n-1)a=n^2a
2
若Sn、Sn+1、Sn+k成等比数列,则:
Sn+1/Sn=(n+1)^2/n^2=Sn+k/Sn+1=(n+k)^2/(n+1)^2
即:(n+1)/n=(n+k)/(n+1)
即:(n+1)^2=n^2+kn
即:kn=2n+1,即:k=2+1/n
k要为正整数,n=1,即:k=3
故:n=1,k=3
an=a1+(n-1)d
即:(a1+(2n-1)d)/(a1+(n-1)d)=(4n-1)/(2n-1)
即:(2n-1)(a1+(2n-1)d)=(4n-1)(a1+(n-1)d)
即:2na1=nd,即:d=2a1=2a
1
{an}通项:an=a1+(n-1)d=a+2(n-1)a=(2n-1)a
Sn=na1+n(n-1)d/2=na+n(n-1)a=n^2a
2
若Sn、Sn+1、Sn+k成等比数列,则:
Sn+1/Sn=(n+1)^2/n^2=Sn+k/Sn+1=(n+k)^2/(n+1)^2
即:(n+1)/n=(n+k)/(n+1)
即:(n+1)^2=n^2+kn
即:kn=2n+1,即:k=2+1/n
k要为正整数,n=1,即:k=3
故:n=1,k=3
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