设数列{an}的前n项和为Sn,且a1=1,Sn=nan-2n(n-1) (1)求a2,a3,a4,并求出数列{an}的通项公式.(2)设数列{1/a

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设数列{an}的前n项和为Sn,且a1=1,Sn=nan-2n(n-1)(1)求a2,a3,a4,并求出数列{an}的通项公式.(2)设数列{1/a设数列{an}的前n项和为Sn,且a1=1,Sn=n

设数列{an}的前n项和为Sn,且a1=1,Sn=nan-2n(n-1) (1)求a2,a3,a4,并求出数列{an}的通项公式.(2)设数列{1/a
设数列{an}的前n项和为Sn,且a1=1,Sn=nan-2n(n-1) (1)求a2,a3,a4,并求出数列{an}的通项公式.(2)设数列{1/a

设数列{an}的前n项和为Sn,且a1=1,Sn=nan-2n(n-1) (1)求a2,a3,a4,并求出数列{an}的通项公式.(2)设数列{1/a
(1)因为a(n+1)=(1+q)an-q*a(n-1),所以有a(n+1)-an=qan-q*a(n-1),又因为bn=a(n+1)-an(n∈N*),所以有bn=q*b(n-1),即bn/b(n-1)=q,(n∈N*),所以{bn}是等比数列;
(2)因为bn=a(n+1)-an(n∈N*),所以b1=a2-a1=1,所以bn=q^(n-1),即a(n+1)-an=q^(n-1),(n∈N*),所以有:an-a(n-1)=q^(n-2),a(n-1)-a(n-2)=q^(n-3),a(n-2)-a(n-3)=q^(n-4),…,a2-a1=1,将上述式子相加得:an-a1=q^(n-2)+q^(n-3)+q^(n-4)+…+1,当q=1时,an=n,当an≠1时,所以an-a1=[1-q^(n-1)]/(1-q),所以an=a1+[1-q^(n-1)]/(1-q),即an=1+[1-q^(n-1)]/(1-q),
(3)当q=1时,有a3=3,a6=6,a9=9,所以a3不是a6与a9的等差中项,当q≠1时,根据(2)可知a3=1+[1-q3)]/(1-q),a6=1+[1-q^6]/(1-q),a9=1+[1-q^9]/(1-q),又因为a3是a6与a9的等差中项,所以1+[1-q^6]/(1-q)+1+[1-q^9]/(1-q)=2[1+(1-q3)/(1-q)},即1-q^6+1-q^9=2-2q3,即q^9+q^6-2q3=0,因为q≠0,所以q^6+q^3-2=0,令q3=t,所以方程变形为t2+t-2=0,解得:t=1(舍去)或者t=-2,即q3=-2,所以q=-3√2


Sn=nan-2n(n-1)
Sn-1=(n-1)a(n-1)-2(n-1)(n-2)

两式相减
得到关于an和an-1的关系式
an-1=an-4
所以 {an}是以1为首相 4为公差的等差数列
an=4n-3
a2=5 a3=9 a4=13

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