1、a(n)=Sn-S(n-1) = -n^2+9n+2-(-(n-1)^2+9(n-1)+2) n>=2
=10-2n
a1 = S1 = 10
an 不是等差數(shù)列
2、當(dāng)n>=5時(shí) |an| = 2n-10
當(dāng)n=5時(shí),Rn = -Sn + 2*S4 = -n^2+9n+46
3、bn = 1/[n*(2n+2)] = (1/2)*[1/n - 1/(n+1)]
Tn = (1/2)*[1/1 - 1/2 + 1/2 -1/3 +……+1/n - 1/(n+1)]
= (1/2)*[1-1/(n+1)]
Tn (1/2)*[1-1/(n+1)] < (n0/32)
1-1/(n+1) < n0/16
要使對(duì)一切非零自然數(shù)n總成立,只要：
1-1/2 < n0/16
n0 > 8
存在n0,且最小為9