由题意可知nan-1=(n-2)an-2,(n-1)an-2=(n-3)an-3,…,5a4=3a3,4a3=2a2,3a2=a1,两边相乘,得n(n+1)an=2a1,由此能够求出an.
【解析】
∵,
∴当n≥2时,
∴Sn-Sn-1=n2an-(n-1)2an-1=an
(n2-1)an=(n-1)2an-1,(n+1)an=(n-1)an-1,
∴nan-1=(n-2)an-2,
(n-1)an-2=(n-3)an-3,
…
5a4=3a3,
4a3=2a2,
3a2=a1,
两边相乘:
3×4×5×…×(n-1)n(n+1)an=1×2×3×…×(n-3)(n-2)(n-1)a1
n(n+1)an=2a1,
∴an==.
故选D.