Question

设Sn是数列{an}的前n项的和,且Sn=2an+n^2-8

设Sn是数列{an}的前n项的和，且Sn=2an+n^2-8
(1)证明数列{an-2n-3}是等比数列
（2）数列{bn}满足bn=1/(an-2n-2)，证明：b1+b2+…+bn<1

Answer 1

(1)证：S(n+1)-Sn=2a(n+1)+(n+1)^2-8-(2an+n^2-8)
==>a(n+1)=2a(n+1)+2n+1-2an
==>a(n+1)-2(n+1)-3=2(an-2n-3)
==>[a(n+1)-2(n+1)-3]/(an-2n-3)=2
所以{an-2n-3}为q=2的等比数列。
(2)证：设cn=1/(an-2n-3)
因为an-2n-2>an-2n-3
==>bn=1/(an-2n-2)<cn=1/(an-2n-3)
而1/(an-2n-3)为q=1/2的等比数列
==>b1+b2+b3+...+bn<c1+c2+c3+...+cn=c1[1-(1/2)^n]/(1-1/2)=2c1[1-(1/2)^n]
S1=a1=2a1+1-8 ==>a1=7, ==>c1=1/(a1-2-3)=1/2
所以b1+b2+b3+...+bn<2*(1/2)*[1-(1/2)^n]=1-(1/2)^n<1
原式得证