s(n) = (n+1)[a(n)+1]/2 - 1.
s(n+1) = (n+2)[a(n+1)+1]/2 - 1,
a(n+1) = s(n+1)-s(n) = [(n+2)a(n+1)-(n+1)a(n)]/2,
na(n+1) = (n+1)a(n),
a(n+1)/(n+1) = a(n)/n,
{a(n)/n}為首項為a(1)/1 = 3,的常數(shù)數(shù)列.
a(n)/n = 3,
a(n) = 3n = 3 + 3(n-1),
{a(n)}是首項為3,公差為3的等差數(shù)列.錯了吧，，S（n＋1）-Sn錯了樓主英明。。。a(n+1) = s(n+1)-s(n) = [(n+2)a(n+1)-(n+1)a(n) + 1]/2,na(n+1) = (n+1)a(n) + 1,a(n+1)/(n+1) = a(n)/n + 1/[n(n+1)] = a(n)/n + 1/n - 1/(n+1),a(n+1)/(n+1) + 1/(n+1) = a(n)/n + 1/n.{a(n)/n + 1/n}為首項為a(1)/1 + 1 = 4,的常數(shù)數(shù)列。a(n)/n + 1/n = 4,a(n) = 4n -1 = 4(n-1) + 3 ,{a(n)}是首項為3，公差為4的等差數(shù)列?，F(xiàn)在是了那，敬請樓主采納~~多謝！其實在那之前我已寫好。。不過給個好評！