Sn=2n^2-n
Sn/(pn+q)=(2n^2-n)/(pn+q)=n(2n-1)/(pn+q)
要使{Sn/(pn+q)}是等差数列,就要使n(2n-1)/(pn+q)满足an+b的形式
即要满足(2n-1)/(pn+q)是常数或者n/(pn+q)是常数
①(2n-1)/(pn+q)是常数
p/q=2/-1=-2
即p=-2q
②n/(pn+q)是常数
q=0,p≠0
Sn=2n^2-n
Sn/(pn+q)=(2n^2-n)/(pn+q)=n(2n-1)/(pn+q)
要使{Sn/(pn+q)}是等差数列,就要使n(2n-1)/(pn+q)满足an+b的形式
即要满足(2n-1)/(pn+q)是常数或者n/(pn+q)是常数
①(2n-1)/(pn+q)是常数
p/q=2/-1=-2
即p=-2q
②n/(pn+q)是常数
q=0,p≠0