Question

数列{an }的前n项和为Sn,已知a1=1,nan-λSn=n(n-1),(n>+2,n属于N+,λ属于R),数列{S

Answer 1

1.

nan-λSn=n(n-1)

an-λ（Sn/n)=n-1

Sn/n=(an-n+1)/λ

S(n-1)/(n-1)=[a(n-1)-n+2]/λ

S(n-2)/(n-2)=[a(n-2)-n+3]/λ

因为2S(n-1)/(n-1)=Sn/n+S(n-2)/(n-2)

所以2[a(n-1)-n+2]/λ=(an-n+1)/λ+[a(n-2)-n+3]/λ

2a(n-1)-2n+4=(an-n+1)+[a(n-2)-n+3]

2a(n-1)=an+a(n-2)

所以an也为等差数列.

设an=1+(n-1)d,sn=n+n(n-1)d/2

nan-λSn=n+n(n-1)d-λ[n+n(n-1)d/2]

=n+n(n-1)d-λn-λn(n-1)d/2

=n(n-1)

n(d-λd/2-1)=d+λ-λd/2-2

d(1-λ/2)-1=0,λ+d(1-λ/2)-2=0

λ=1,d=2.

2.

由上得an=1+(n-1)d=2n-1

bn=(an+λ)n=(2n-1+1)n=2n^2

Tn=2(n-1)n(2n-1)/6=(n-1)n(2n-1)/3