- 数列{an}的前n项和为Sn,当n>=1时,S(n+1)是?
- 数列{an}的前n项和为Sn,当n>=1时,S(n+1)是a(n+1)与
S(n+1)+2的等比中项.
1.求证:当n>=时,1/Sn-1/S(n+1)=1/2
2.设a1=-1,求Sn的表达式
3.设a1=-1,且{n/[(pn+q)Sn]}是等差数列{pq不等于0},求证:p/q是常数.
- 1.S(n+1)是a(n+1)与 S(n+1)+2的等比中项.
[S(n+1)]^2=a(n+1)[S(n+1)+2]=[s(n+1)-s(n),S(n+1)+2]
=[s(n+1)]^2+2s(n+1)-s(n)s(n+1)-2s(n),
2s(n+1)-s(n)s(n+1)-2s(n)=0,
1/[s(n+1)-1/[S(n)]=-1/2,知结论成立。
2.s1=a1=-1,1/Sn=-1-1/2*(n-1)=-(n+1)/2,Sn=-2/(n+1).
3.设f(n)=n/[(pn+q)Sn],则
f(3)+f(1)=2f(2),
即 -6/(3p+q)-1/(p+q)=-6/(2p+q),
两边乘以-(3p+q)(p+q)(2p+q),得
6(p+q)(2p+q)+(3p+q)(2p+q)=6(3p+q)(p+q)
(3p+q)(2p+q)=6p(p+q),
6p^2+5pq+q^2=6p^2+6pq,
q^2=pq,
p/q=1.