已知数列{an}的前n项和为Sn,满足Sn=2an+n^2-4n(n=1,2,3…)(1) 写出数列{an}的前3项a1,a2,a3(2) 求证,数列{an-2n+1}为等比数列(3) 求Sn
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![已知数列{an}的前n项和为Sn,满足Sn=2an+n^2-4n(n=1,2,3…)(1) 写出数列{an}的前3项a1,a2,a3(2) 求证,数列{an-2n+1}为等比数列(3) 求Sn](/uploads/image/z/4481485-61-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%BB%A1%E8%B6%B3Sn%EF%BC%9D2an%EF%BC%8Bn%EF%BC%BE2%EF%BC%8D4n%EF%BC%88n%EF%BC%9D1%2C2%2C3%E2%80%A6%EF%BC%89%EF%BC%881%EF%BC%89+%E5%86%99%E5%87%BA%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8D3%E9%A1%B9a1%2Ca2%2Ca3%EF%BC%882%EF%BC%89+%E6%B1%82%E8%AF%81%2C%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BC%8D2n%EF%BC%8B1%EF%BD%9D%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%883%EF%BC%89+%E6%B1%82Sn)
已知数列{an}的前n项和为Sn,满足Sn=2an+n^2-4n(n=1,2,3…)(1) 写出数列{an}的前3项a1,a2,a3(2) 求证,数列{an-2n+1}为等比数列(3) 求Sn
已知数列{an}的前n项和为Sn,满足Sn=2an+n^2-4n(n=1,2,3…)
(1) 写出数列{an}的前3项a1,a2,a3
(2) 求证,数列{an-2n+1}为等比数列
(3) 求Sn
已知数列{an}的前n项和为Sn,满足Sn=2an+n^2-4n(n=1,2,3…)(1) 写出数列{an}的前3项a1,a2,a3(2) 求证,数列{an-2n+1}为等比数列(3) 求Sn
用(n-1)代入n得出S(n-1)两式相减:an=Sn-S(n-1)=.化简即可,时间紧没算,但思路是这样
(1)a1=2a1+1-4 a1=3
a2+a1=2a2+4-8 a2=7
a3+a2+a1=2a3+9-12 a3=13
(2)
当n>1(n∈N+)时,
Sn=2an+n^2-4n ①
S(n-1)=2a(n-1)+(n-1)^2-4(n-1)②
①-②,得
an=2an-2a(n-1)+2n-5<...
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(1)a1=2a1+1-4 a1=3
a2+a1=2a2+4-8 a2=7
a3+a2+a1=2a3+9-12 a3=13
(2)
当n>1(n∈N+)时,
Sn=2an+n^2-4n ①
S(n-1)=2a(n-1)+(n-1)^2-4(n-1)②
①-②,得
an=2an-2a(n-1)+2n-5
移项,得an=2a(n-1)-2n+5
配凑:an-2n+1=2a(n-1)-4n+6
an-2n+1=2a(n-1)-4(n-1)+2
an-2n+1=2(a(n-1)-2(n-1)+1)
a2-2*2+1=4
a1-2*1+1=2
所以,数列{an-2n+1}是以2为首项,2为公比的等比数列
(3)an-2n+1=2^n
所以an=2^n+2n-1
Sn=a1+a2+a3+...+an
=2^1+2*1-1+2^2+2*2-1+2^3+2*3-1+...+2^n+2*n-1
=2(1-2^n)/(1-2)+(2+2n)n/2-n
=2^(n+1)+n^2-2
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