Question

设数列{an}的各项都为正数,其前n项和为sn,已知对任意n,sn是an的平方和an的等差

Answer 1

(1) (an+2)/2=根号下2Sn

所以8Sn=(an+2)^2

n=1,S1=a1.8a1=(a1+2)^2,得a1=2

n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6

n=3,8S3=(a1+2)^2,8(a1+a2+a3)=(a3+2)^2,得a3=10

(2) 8Sn=(an+2)^2

当n≥2时,8S(n-1)=[a(n-1)+2]^2

两式相减得8an=(an+2)^2-[a(n-1)+2]^2

(an)^2+4an+4-[a(n-1)]^2-4a(n-1)-4=8an

(an)^2-[a(n-1)]^2-4an-4a(n-1)=0

[an+a(n-1)][an-a(n-1)]-4[an+a(n-1)]=0

[an+a(n-1)][an-a(n-1)-4]=0

∵ {an}是正数组成的数列,∴an>0,a(n-1)>0

∴ an-a(n-1)=4

∴ {an}是等差数列,首项为a1,公差为4.

∴ an=a1+(n-1)d=2+4(n-1)=4n-2