数列{An}中,a1=2,a (n+1)=4an-3n+1,n为N*1,证明:数列{an - n}是等比数列2,求数列{an}前n项和Sn3,证明不等式S(n+1)< = 4Sn,对任意n为N* 成立
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/14 01:40:01
![数列{An}中,a1=2,a (n+1)=4an-3n+1,n为N*1,证明:数列{an - n}是等比数列2,求数列{an}前n项和Sn3,证明不等式S(n+1)< = 4Sn,对任意n为N* 成立](/uploads/image/z/2717321-41-1.jpg?t=%E6%95%B0%E5%88%97%7BAn%7D%E4%B8%AD%2Ca1%3D2%2Ca+%28n%2B1%29%3D4an-3n%2B1%2Cn%E4%B8%BAN%2A1%2C%E8%AF%81%E6%98%8E%EF%BC%9A%E6%95%B0%E5%88%97%7Ban+-+n%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%972%2C%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E5%89%8Dn%E9%A1%B9%E5%92%8CSn3%2C%E8%AF%81%E6%98%8E%E4%B8%8D%E7%AD%89%E5%BC%8FS%EF%BC%88n%2B1%29%3C+%3D+4Sn%2C%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E4%B8%BAN%2A+%E6%88%90%E7%AB%8B)
xՒJ@_enb$xJAPRkE7/^DoH/&$lvf~;0sAVsȳ@_|PjS>l{cb7:l5 ?,|S.'TӤr+D
bQ/|["_PSZ
qFLw҇S(֚^EI,RHTo 3^z
("HD:1%Q
数列{An}中,a1=2,a (n+1)=4an-3n+1,n为N*1,证明:数列{an - n}是等比数列2,求数列{an}前n项和Sn3,证明不等式S(n+1)< = 4Sn,对任意n为N* 成立
数列{An}中,a1=2,a (n+1)=4an-3n+1,n为N*
1,证明:数列{an - n}是等比数列
2,求数列{an}前n项和Sn
3,证明不等式S(n+1)< = 4Sn,对任意n为N* 成立
数列{An}中,a1=2,a (n+1)=4an-3n+1,n为N*1,证明:数列{an - n}是等比数列2,求数列{an}前n项和Sn3,证明不等式S(n+1)< = 4Sn,对任意n为N* 成立
1.
a (n+1)=4an-3n+1
=>
a(n+1) - (n+1) = 4(an -n)
{an - n}是等比数列
2.
an-n = 4^(n-1)*(a1-1)=4^(n-1)
=>
an=4^(n-1) + n
Sn = (1+4+16+……+4^(n-1))+(1+2+3+……+n)
=(4^n - 1)/(4-1) + n(n+1)/2
=(4^n - 1)/3 + n(n+1)/2
3.
S(n+1) - Sn = a(n+1) = 4^n + n + 1
S(n+1)< = 4Sn
S(n+1)-Sn
4^n + n+1