数列an中,a1=1/4 ,当n>=2时,有(3n^2-2n-1)an=a1+a2+a3+.+a(n-1)(1).求an(2).求前n项和Sn

数列an中,a1=1/4 ,当n>=2时,有(3n^2-2n-1)an=a1+a2+a3+.+a(n-1)(1).求an(2).求前n项和Sn
数列an中,a1=1/4 ,当n>=2时,有(3n^2-2n-1)an=a1+a2+a3+.+a(n-1)
(1).求an
(2).求前n项和Sn

数列an中,a1=1/4 ,当n>=2时,有(3n^2-2n-1)an=a1+a2+a3+.+a(n-1)(1).求an(2).求前n项和Sn
(3n^2-2n-1)an=a1+a2+a3+.+a(n-1)
[3(n-1)^2-2(n-1)-1]an-1=a1+a2+a3+.+a(n-2)
两式相减
(3n^2-2n-1)an-[3(n-1)^2-2(n-1)-1]a(n-1)=a(n-1)
(3n^2-2n-1)an=[3n^2-6n+3-2n+2-1+1]a(n-1)
(3n^2-2n-1)an=[3n^2-8n+5]a(n-1)
(3n+1)(n-1)an=[3n-5][n-1]a(n-1)
(3n+1)an=(3n-5)a(n-1)
an/a(n-1)=(3n-5)/(3n+1)
an/a(n-1)=[3(n-2)+1]/(3n+1)
所以,有
a(n-1)/a(n-2)=[3(n-3)+1]/[3(n-1)+1]
.
a4/a3=[3x2+1]/[3x4-1]
a3/a2=[3x1+1]/[3x3-1]
相乘得到
an/a2 = [3x2+1][3x1+1]/(3n+1)[3(n-1)+1](注：观察可以得到分母只剩下开始的两项,而分子只剩下最后的两项)
整理得到：
an=28a2/(3n+1)[3(n-1)+1]
由条件
(12-4-1)a2=1/4
a2=1/28
代入上式
an = 1/(3n+1)[3(n-1)+1]
当n=1时an = 1/4,在通项公式内所以
an = 1/(3n+1)[3(n-1)+1]
第一问OK
an = 1/(3n+1)[3(n-1)+1]
=1/3[1/[3(n-1)+1] - 1/(3n+1)]
所以
Sn = a1+a2+...+an
=1/3[1-1/4+1/4-1/7+.+1/[3(n-1)+1] - 1/(3n+1)]
=1/3[1-1/(3n+1)]
=n/(3n+1)
第二问OK
再验证一下吧,整体上做得比较顺利,应该没大问题

(3n^2-2n-1)an=a1+a2+a3+......+a(n-1)
[3(n-1)^2-2(n-1)-1]an-1=a1+a2+a3+......+a(n-2)
两式相减
(3n^2-2n-1)an-[3(n-1)^2-2(n-1)-1]a(n-1)=a(n-1)
(3n^2-2n-1)an=[3n^2-6n+3-2n+2-1+1]a(n-1)
(3n...

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(3n^2-2n-1)an=a1+a2+a3+......+a(n-1)
[3(n-1)^2-2(n-1)-1]an-1=a1+a2+a3+......+a(n-2)
两式相减
(3n^2-2n-1)an-[3(n-1)^2-2(n-1)-1]a(n-1)=a(n-1)
(3n^2-2n-1)an=[3n^2-6n+3-2n+2-1+1]a(n-1)
(3n^2-2n-1)an=[3n^2-8n+5]a(n-1)
(3n+1)(n-1)an=[3n-5][n-1]a(n-1)
(3n+1)an=(3n-5)a(n-1)
an/a(n-1)=(3n-5)/(3n+1)
an/a(n-1)=[3(n-2)+1]/(3n+1)
所以，有
a(n-1)/a(n-2)=[3(n-3)+1]/[3(n-1)+1]
....
a4/a3=[3x2+1]/[3x4-1]
a3/a2=[3x1+1]/[3x3-1]
相乘得到
an/a2 = [3x2+1][3x1+1]/(3n+1)[3(n-1)+1](注：观察可以得到分母只剩下开始的两项，而分子只剩下最后的两项)
整理得到：
an=28a2/(3n+1)[3(n-1)+1]
由条件
(12-4-1)a2=1/4
a2=1/28
代入上式
an = 1/(3n+1)[3(n-1)+1]
当n=1时an = 1/4，在通项公式内所以
an = 1/(3n+1)[3(n-1)+1]

第一问OK

an = 1/(3n+1)[3(n-1)+1]
=1/3[1/[3(n-1)+1] - 1/(3n+1)]
所以
Sn = a1+a2+...+an
=1/3[1-1/4+1/4-1/7+....+1/[3(n-1)+1] - 1/(3n+1)]
=1/3[1-1/(3n+1)]
=n/(3n+1)
第二问OK

再验证一下吧，整体上做得比较顺利，应该没大问题

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