jiu00
幼苗
共回答了20个问题采纳率:85% 举报
an=sn-sn-1=n^2/2+11n/2-[(n-1)^2/2+11(n-1)/2]=(n^2-(n-1)^2)/2+11/2=n+5
b(n+2)-2b(n+1)+bn=0
b(n+2)-b(n+1)=b(n+1)-bn
数列{bn}为等差数列.
b3=b1+2d=11,(1)
b1+b2+..+b9=9b1+9(9-1)d/2=153(2)
(2)-(1)*9,得
b1=5,d=3
bn=b1+(n-1)d=5+(n-1)*3=3n+2
cn=3/[(2an-11)(2bn-1)]=3/[(2n+5)-11)*(2(3n+2)-1)]=1/[(2n-1)(2n+1)]=(1/2)(1/(2n-1)-1/(2n+1))
Tn=(1/2){[1/(2*1-1)-1(2*1+1)]+[1/(2*2-1)-1/(2*2+1)]+.+[1/(2n-1)-1/(2n+1)]}
=(1/2){[1+1/3+...+1/(2n-1)]-[1/3+1/5+.+1/(2n+1)]}
= (1/2)[1-1/(2n+1)]
=n/(2n+1)=1/(2+1/n)
Tn>k/57
则1/(2+1/n)>k/57=1/(57/K)
(2+1/n)3
所以k
1年前
10