Question

高中数列第24题求学霸解答

Answer 1

(1)由已知条件知[Sn-(n^2+n+1)][Sn+1]=0因为an是正项数列，Sn>0,Sn不等于-1所以Sn=n^2+n+1
an=Sn-Sn-1=(n^2+n+1)-[(n-1)^2+(n-1)+1]=2n
(2)bn=(n+1)/(n+2)^2(2n)^2
=1/4{n+1)/(n+2)^2(n)^2}
=1/4{1/4[1/n^2-1/(n+2)^2]}
=1/16[1/n^2-1/(n+2)^2]
bn-1=1/16[1/(n-1)^2-1/(n+1)^2]
bn-2=1/16[1/(n-2)^2-1/(n)^2]
bn-3=1/16[1/(n-3)^2-1/(n-1)^2]
bn-4=1/16[1/(n-4)^2-1/(n-2)^2]
……启告
b3=1/16[1/3^2-1/5^2]
b2=1/16[1/2^2-1/4^2]
b1=1/16[1-1/3^2]=1/18
T1=b1=1/18
Tn-T1=1/16[-1/(n+2)^2-1/(n+1)^2+1/9+1/4]+1/扒哪18-1/悄此明18=1/16*(1/9+1/4)-1/16[-1/(n+2)^2-1/(n+1)^2]
<1/16*(1/9+1/4)=13/576
所以n≥2时，Tn-T1<13/576

Answer 2

第一辩伍迅问 分解因式得sn =n^2+n +1或sn=-1,因为sn >0 即为n^2+n+1 然后橘改求出an =2n n>=2时携此a1=3

追答

第二问 化简bn =1/16乘（1/n^2 -1/(n+2)^2 ）n>=2时 b1=2/81然后求其和Tn-T1=1/16乘（1/4+1/9-1/（n+1)^2 -1/(n+2)^2）<1/16乘1/4-1/9=13/576