已知数列{an}的首项a1=1⼀2,其前n项和为sn=n^2an(n≥1)求通项

2024-12-19 12:18:31
推荐回答(1个)
回答1:

Sn=n^2an
=n^2(Sn-S(n-1))
(n^2-1)Sn = n^2S(n-1)
(n-1)(n+1)Sn = n^2S(n-1)
[(n+1)/n]Sn = [n/(n-1)]S(n-1)
[(n+1)/n]Sn =nS1
= n/2

Sn = n^2/[2(n+1)]
an = Sn- S(n-1)
= n^2/(n+1) - (n-1)^2/n
= [n^3- (n-1)^2.(n+1)]/[n(n+1)]
= (n^2 +n-1)[n(n+1)]