Question

数列{an}的前n项和为Sn，已知a1=1/2，Sn=n^2an-n(n-1)，n=1,2,… (

Answer 1

a(1)=s(1)=1/2,

s(n) = n^2a(n)-n(n-1),

s(n+1) = (n+1)^2a(n+1) - (n+1)n = (n+1)^2[s(n+1)-s(n)] - (n+1)n

= (n+1)^2s(n+1) - s(n)(n+1)^2 - (n+1)n,

n(n+2)s(n+1) = (n+1)^2s(n) + (n+1)n,

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