设数列{an}的前n项和为Sn,a1+2a2+3a3.+nan=(n-1)Sn+2n,设数列{an}的前n项和为Sn,a1+2a2+3a3.+nan=(n-1)Sn+2n1)求Sn+2为等比.2)抽取an中的第1,4,7..3n-2项,余下各项顺序不变,组成新数列{bn},{bn}前n项和为Tn,证明12/5
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![设数列{an}的前n项和为Sn,a1+2a2+3a3.+nan=(n-1)Sn+2n,设数列{an}的前n项和为Sn,a1+2a2+3a3.+nan=(n-1)Sn+2n1)求Sn+2为等比.2)抽取an中的第1,4,7..3n-2项,余下各项顺序不变,组成新数列{bn},{bn}前n项和为Tn,证明12/5](/uploads/image/z/1741021-61-1.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Ca1%2B2a2%2B3a3.%2Bnan%3D%28n-1%29Sn%2B2n%2C%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Ca1%2B2a2%2B3a3.%2Bnan%3D%28n-1%29Sn%2B2n1%EF%BC%89%E6%B1%82Sn%2B2%E4%B8%BA%E7%AD%89%E6%AF%94.2%29%E6%8A%BD%E5%8F%96an%E4%B8%AD%E7%9A%84%E7%AC%AC1%2C4%2C7..3n-2%E9%A1%B9%2C%E4%BD%99%E4%B8%8B%E5%90%84%E9%A1%B9%E9%A1%BA%E5%BA%8F%E4%B8%8D%E5%8F%98%2C%E7%BB%84%E6%88%90%E6%96%B0%E6%95%B0%E5%88%97%7Bbn%7D%2C%7Bbn%7D%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E8%AF%81%E6%98%8E12%2F5)
设数列{an}的前n项和为Sn,a1+2a2+3a3.+nan=(n-1)Sn+2n,设数列{an}的前n项和为Sn,a1+2a2+3a3.+nan=(n-1)Sn+2n1)求Sn+2为等比.2)抽取an中的第1,4,7..3n-2项,余下各项顺序不变,组成新数列{bn},{bn}前n项和为Tn,证明12/5 设数列{an}的前n项和为Sn,a1+2a2+3a3.+nan=(n-1)Sn+2n,设数列{an}的前n项和为Sn,a1+2a2+3a3.+nan=(n-1)Sn+2n1)求Sn+2为等比.2)抽取an中的第1,4,7..3n-2项,余下各项顺序不变,组成新数列{bn},{bn}前n项和为Tn,证明12/5 证明12/5 2B啊 本想帮你的
设数列{an}的前n项和为Sn,a1+2a2+3a3.+nan=(n-1)Sn+2n,
设数列{an}的前n项和为Sn,a1+2a2+3a3.+nan=(n-1)Sn+2n
1)求Sn+2为等比.2)抽取an中的第1,4,7..3n-2项,余下各项顺序不变,组成新数列{bn},{bn}前n项和为Tn,证明12/5
1)
a1=2,
a1+2a2=a1+a2+4
a2=4
a1+2a2+3a3=2(a1+a2+a3)+6
10+3a3=12+2a3+6
a3=8
(a1+a2+a3+.+an)+(a2+a3+.+an)+.+an=sn+(sn-s1)+(sn-s2)+.+(sn-s(n-1))
n*sn-(s1+s2+...+s(n-1))=(n-1)sn+2n
sn=(s1+s2+.+s(n-1))+2n
s(n+1)=(s1+s2+.+sn))+2(n+1)
a(n+1)=sn+2
an=s(n-1)+2
a(n+1)-an=an
a(n+1)=2an
{an}是等比数列
an=2^n
sn=2(2^n-1)=2^(n+1)-2
sn+2=2^(n+1)
[s(n+1)+2]/[sn+2]=2为常数
所以Sn+2 等比数列
2)
an=2^n
抽取后得到bn
2^2,2^3,2^5,2^6,2^8,2^9```
显然
奇项是以4为首,8为公比的GP
偶项是以8为首,8为公比的GP
当n=2k-1,k∈N+
Tn=b1+b3+```+b2k-1+b2+b4+```+b2k-2
=2^2+2^5+```+2^3k-1+2^3+2^6+```2^3k-3
=(10/7)*2^(3k-1)-12/7=5/7*8^k-12/7 (两等比求和再求和)
当n=2k,k∈N+
Tn=b1+b3+```+b2k-1+b2+b4+```+b2k
=(12/7)*2^(3k)-12/7=12/7*8^k-12/7
下对n奇偶性分类分别证明不等式即可