Solution: Because S2=3 and 2Sn=n+nan, that is, 2S2 = 2+2A2, so A2 = S2- 1 = 2, then A 1 = 1

When n≥2, 2an = 2sn-2s (n-1) = nan-(n- 1)a(n- 1), that is, (n-1) a (n-/kloc-0

Divide both sides by (n- 1)(n-2) to get an/(n- 1)= a(n- 1)/(n-2), where an/(n-1) = a (n-).

So an = {( 1, n =1); (2(n- 1), n≥2). (Piecewise function form)

BN = 2 (n- 1) (process omitted)

I have something to do. I'll finish the following when I get back. I'm sorry