已知数列{an}的前n项和为Sn,且有Sn=n^2/2+11n/2,数列{bn}满足bn+2-2bn+1+bn=(n∈N+),且b3=11,前9项和为153(1)求数列{an}、{bn}的通项公式;(2)设cn=3/(2an-11)(2bn-1),数列{cn}的前n项和为Tn,求
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![已知数列{an}的前n项和为Sn,且有Sn=n^2/2+11n/2,数列{bn}满足bn+2-2bn+1+bn=(n∈N+),且b3=11,前9项和为153(1)求数列{an}、{bn}的通项公式;(2)设cn=3/(2an-11)(2bn-1),数列{cn}的前n项和为Tn,求](/uploads/image/z/4531294-46-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E6%9C%89Sn%3Dn%5E2%2F2%2B11n%2F2%2C%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E6%BB%A1%E8%B6%B3bn%2B2-2bn%2B1%2Bbn%3D%EF%BC%88n%E2%88%88N%2B%EF%BC%89%2C%E4%B8%94b3%3D11%2C%E5%89%8D9%E9%A1%B9%E5%92%8C%E4%B8%BA153%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E3%80%81%EF%BD%9Bbn%EF%BD%9D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%EF%BC%882%EF%BC%89%E8%AE%BEcn%3D3%2F%282an-11%29%282bn-1%29%2C%E6%95%B0%E5%88%97%EF%BD%9Bcn%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E6%B1%82)
已知数列{an}的前n项和为Sn,且有Sn=n^2/2+11n/2,数列{bn}满足bn+2-2bn+1+bn=(n∈N+),且b3=11,前9项和为153(1)求数列{an}、{bn}的通项公式;(2)设cn=3/(2an-11)(2bn-1),数列{cn}的前n项和为Tn,求
已知数列{an}的前n项和为Sn,且有Sn=n^2/2+11n/2,数列{bn}满足bn+2-2bn+1+bn=(n∈N+),且b3=11,前9项和为153
(1)求数列{an}、{bn}的通项公式;
(2)设cn=3/(2an-11)(2bn-1),数列{cn}的前n项和为Tn,求使不等式Tn>k/57对一切n∈N+都成立的最大正整数k的值;
主要是最后一小题的求k的值,Tn我已经算出来了
已知数列{an}的前n项和为Sn,且有Sn=n^2/2+11n/2,数列{bn}满足bn+2-2bn+1+bn=(n∈N+),且b3=11,前9项和为153(1)求数列{an}、{bn}的通项公式;(2)设cn=3/(2an-11)(2bn-1),数列{cn}的前n项和为Tn,求
1、
a1=S1=1/2+11/2=6
Sn=n²/2 +11n/2
S(n-1)=(n-1)²/2 +11(n-1)/2
S(n-1)-Sn=an=n²/2 +11n/2 -(n-1)²/2 -11(n-1)/2=n+5
n=1时,a1=1+5=6,同样满足.
数列{an}的通项公式为an=n+5.
b(n+2)-2b(n+1)+bn=0
b(n+2)-b(n+1)=b(n+1)-bn
数列{bn}是等差数列,设公差为d.
S9'=9b1+36d=9(b1+4d)=9b5=153
b5=17
b5-b3=2d=17-11=6 d=3
b1=b3-2d=11-6=5
bn=5+3(n-1)=3n+2
数列{bn}的通项公式为bn=3n+2.
2、
cn=3/[(2an-11)(2bn-1)]=2/[2(n+5)-11][2(3n+2)-1]=2/[(2n-1)(6n+3)]
=(2/3)/[(2n-1)(2n+1)]=(1/3)[1/(2n-1)-1/(2n+1)]
Tn=c1+c2+...+cn
=(1/3)[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=(1/3)[1-1/(2n+1)]
=2n/[3×(2n+1)]
=2/[3×(2+ 1/n)]
随n增大,1/n减小,2+ 1/n减小,3×(2+ 1/n)减小,2/[3×(2+ 1/n)]增大,当n=1时,Tn取得最小值Tmin=2/[3×(2+1)]=2/9
Tn>k/57对于一切n∈N+都成立,则当Tn取得最小值时等式同样成立.
2/9>k/57
k