Question

设sn为数列的前n项和,sn=kn^2+n,其中k是常数【要详细过程

Answer 1

1.

令n=1,a1=S1=k×1^2 +1=k+1

n≥2时,an=Sn-S(n-1)=kn^2 +n-[k(n-1)^2+(n-1)]=k(2n-1)+1=2kn-k+1

n=1时,a1=2k×1-k+1=k+1,同样满足

综上,得

a1=k+1

an=2kn-k+1

2.

am,a2m,a4m成等比数列,则a(2m)^2=am·a(4m)

[2k·(2m)-k+1]^2=(2km-k+1)(2k·4m-k+1)

整理,得km(k-1)=0

m为正整数,因此只有k=0或k=1

k=0时,an=1,数列是以1为首项,1为公比的等比数列,满足题意

k=1时,an=2n-1+1=2n

am·a(4m)=2m·2·(4m)=16m^2

a(2m)^2=(2·2m)^2=16m^2

a(2m)^2=am·a(4m),am,a(2m),a(4m)成等比数列,满足题意

综上,得k=0或k=1