正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn
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![正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn](/uploads/image/z/11458406-38-6.jpg?t=%E6%AD%A3%E6%95%B0%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8Fn%3DN%2B%E9%83%BD%E6%9C%892sn%2Bnan%3D6n%2B3%2C%E6%B1%82a1%2Ca2%2Ca3%3B%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%E4%BB%A4bn%3D%28an-2%29%28n%2B1%29%2Cbn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E6%B1%82%E8%AF%81Tn)
正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn
正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;
令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn<9/2
正数数列an的前n项和为sn,且对任意n=N+都有2sn+nan=6n+3,求a1,a2,a3;求an的通项公式;令bn=(an-2)(n+1),bn的前n项和为Tn,求证Tn
(1)
2Sn+nan=6n+3
n=1
3a1=9
a1=3
n=2
2(a1+a2)+2a2=15
4a2=9
a2=9/4
n=3
2(a1+a2+a3)+3a3=21
5a3=21/2
a3=21/10
2Sn+nan=6n+3 (1)
2S(n-1)+(n-1)a(n-1)=6(n-1)+3 (2)
(1)-(2)
2an+nan-(n-1)a(n-1)=6
(2+n)[an -2] = (n-1)( a(n-1) - 2)
(an-2)/(a(n-1)-2) = (n-1)/(n+2)
(an-2)/(a1 -2 ) = 6/[(n+2)(n+1)n]
an = 2+ 6/[(n+2)(n+1)n]
(2)
bn = (an-2)(n+1)
= 6/[n(n+2)]
= 3[ 1/n - 1/(n+2) ]
Tn =b1+b2+..+bn
= 3[ 1+1/2 - 1/(n+1)-1/(n+2) ]
< 3(3/2)
=9/2
(1)
2Sn+nan=6n+3
n=1
3a1=9
a1=3
n=2