(1)
令n=1
a1=S1=3-1=2
Sn=3n²-n
Sn-1=3(n-1)²-(n-1)
an=Sn-Sn-1=3n²-n-3(n-1)²+(n-1)=6n-4
n=1時(shí),a1=6-4=2,同樣滿足,數(shù)列{an}的通項(xiàng)公式為an=6n-4
a(n+1)=6(n+1)-4=6n+2
a(n+1)-an=6(n+1)-4-6n+4=6
bn=1/[√an+√a(n+1)]=[√a(n+1)-√an]/[a(n+1)-an]=[√a(n+1)-√an]/6
Tn=b1+b2+...+bn
=[√a2-√a1+√a3-√a2+...+√a(n+1)-√an]/6
=[√a(n+1)-√a1]/6
=[√(6n+2)-√2]/6
=√2[√(3n+1)-1]/6
(2)
Sn/n-1=(3n²-n)/n-1=3n-1-1=3n-2=(1/2)(6n-4)=(1/2)an
an指數(shù)是1,函數(shù)Sn/(n-1)=(1/2)an是一次函數(shù),圖像是一條直線,即點(diǎn)Pn（an,Sn/n-1)(n=1,2,...)在同一條直線上,直線方程為Sn/(n-1)=(1/2)an