Question

数列问题

设数列{an}的通项为an=kn+2，其前n项和为Sn，如果数列{S2n/Sn}是常数列，求常数k的值。

Answer 1

an是以2为首项，k为公差的等差数列
所以 Sn=2n+kn(1+n)/2
所以S2n = k*2n*(2n+1)/2 + 2*2n= kn(2n+1) + 4n
S2n/Sn = (2kn+k+4)/(kn/2+k/2+2)
因为S2n/Sn为 常数
所以 2kn/(kn/2)=(k+4)/(k/2+2)
可以得到 k=-4

Answer 2

S2n=(a1+a2n)*2n/2
Sn=(a1+an)*n/2
S2n/Sn
=2(a1+a2n)/(a1+an)
=2(k+2+2nk+2)/(k+2+kn+2)
=2(2nk+k+4)/(kn+k+4)
=2(2nk+2k+8-k-4)/(kn+k+4)
=2[(2nk+2k+8)/(kn+k+4)-(k+4)/(kn+k+4)]
=2[2-(k+4)/(kn+k+4)]
是常数则只有(k+4)/(kn+k+4)=0
k+4=0
k=-4