(1)
n=1時,a1=S1=k+1+r
n≥2時,Sn=kn^2+n+r S(n-1)=k(n-1)^2+(n-1)+r
an=Sn-S(n-1)=2kn-k+1
an-a(n-1)=2k
要數(shù)列是等差數(shù)列,a1同樣滿足通項公式
a1=2k-k+1=k+1,又a1=k+1+r,因此
k+1=k+1+r
r=0
(2)
r=0 Sn=kn^2+n
n=1時,a1=S1=k+1
n≥2時,S(n-1)=k(n-1)^2+(n-1)
an=Sn-S(n-1)=2kn-k+1
a(2m),a(4m),a(8m)成等比數(shù)列
[a(4m)]^2=a(2m)a(8m)
[(2k)(4m) -k+1]^2=[(2k)(2m)-k+1][(2k)(8m)-k+1]
整理,得
4km(k-1)=0
m為任意正整數(shù),要等式成立,k=0或k=1