等差数列{an}中,a1=1,前n项和Sn满足条件S(2n)/Sn=(4n+2)/(n+1)(n=1,2,……)1)求数列{an}的通项公式2)记bn=anp^an(p>0),求{bn}的前n项和Tn┓ ┓ ┣━━━┓┏┣┏┣━┓┏┻━━━┓┃┃ ┃ ┃
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![等差数列{an}中,a1=1,前n项和Sn满足条件S(2n)/Sn=(4n+2)/(n+1)(n=1,2,……)1)求数列{an}的通项公式2)记bn=anp^an(p>0),求{bn}的前n项和Tn┓ ┓ ┣━━━┓┏┣┏┣━┓┏┻━━━┓┃┃ ┃ ┃](/uploads/image/z/1770628-4-8.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D1%2C%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3%E6%9D%A1%E4%BB%B6S%EF%BC%882n%EF%BC%89%2FSn%3D%EF%BC%884n%2B2%EF%BC%89%2F%EF%BC%88n%2B1%EF%BC%89%EF%BC%88n%3D1%2C2%2C%E2%80%A6%E2%80%A6%EF%BC%891%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F2%EF%BC%89%E8%AE%B0bn%3Danp%5Ean%28p%3E0%29%2C%E6%B1%82%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn%E2%94%93+%E2%94%93+%E2%94%A3%E2%94%81%E2%94%81%E2%94%81%E2%94%93%E2%94%8F%E2%94%A3%E2%94%8F%E2%94%A3%E2%94%81%E2%94%93%E2%94%8F%E2%94%BB%E2%94%81%E2%94%81%E2%94%81%E2%94%93%E2%94%83%E2%94%83+%E2%94%83+%E2%94%83)
等差数列{an}中,a1=1,前n项和Sn满足条件S(2n)/Sn=(4n+2)/(n+1)(n=1,2,……)1)求数列{an}的通项公式2)记bn=anp^an(p>0),求{bn}的前n项和Tn┓ ┓ ┣━━━┓┏┣┏┣━┓┏┻━━━┓┃┃ ┃ ┃
等差数列{an}中,a1=1,前n项和Sn满足条件S(2n)/Sn=(4n+2)/(n+1)(n=1,2,……)
1)求数列{an}的通项公式
2)记bn=anp^an(p>0),求{bn}的前n项和Tn
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等差数列{an}中,a1=1,前n项和Sn满足条件S(2n)/Sn=(4n+2)/(n+1)(n=1,2,……)1)求数列{an}的通项公式2)记bn=anp^an(p>0),求{bn}的前n项和Tn┓ ┓ ┣━━━┓┏┣┏┣━┓┏┻━━━┓┃┃ ┃ ┃
1)因为Sn=na1+n(n-1)d/2=n+n(n-1)d/2,S2n=2n+2n(2n-1)d/2,
S(2n)/Sn=(4n+2)/(n+1),所以d=1,所以Sn=n+n(n-1)/2
2)an=n,所以bn=n*p^n,
bn=p*b(n-1)+p^n
b(n-1)=p*b(n-2)+p^(n-1)
b(n-2)=p*b(n-3)+p^(n-2)
.
b2=p*b1+p^2
相加起来得到:Tn-b1=p*[T(n-1)]+p^2[1-p^(n-1)]/(1-p)
Tn-b1=p*[Tn-bn]+p^2[1-p^(n-1)]/(1-p)
Tn(1-p)=b1-bn*p+p^2[1-p^(n-1)]/(1-p)
Tn={p-[n*p^n]*p+p^2[1-p^(n-1)]/(1-p)}/(1-p)
化简得到:Tn=[1-p^(n+1)+n*(p-1)*p^(n+1)]/(1-p)^2