Question

数列{an}的前n项和为Sn,已知a1=1,nan-λSn=n(n-1)(n≥2,n∈N+,λ∈R),数列{Sn/n}为

Answer 1

nan-λSn=n(n-1)

n=1

a1-λa1=0

λ=1

Sn = n(an-n+1) (1)

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

(1)-(2)

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

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

an = a(n-1) +2

an -a(n-1) =2

an -a1 = 2(n-1)

an = 2n-1

bn = (an+λ)^n

= (2n)^n 真的是n次方?