已知等差数列{an}的前n项和是Sn,且a1=2,S5=30

(1)
an=a1+(n-1)d= 2+(n-1)d
S5 = (2+2d)5 = 30
d=2
an = 2+(n-1)2 = 2n
(2)
bn=2^(an-1)
= 2^(2n-1)
Tn =b1+b2+...+bn
= 2(2^(2n) -1) / (4-1)
= (2/3)(2^(2n) -1)
(3)
cn =1/Sn
= 1/ [n(n+1)]
= 1/n -1/(n+1)
Rn = c1+c2+...+cn
= 1-1/(n+1)
= n/(n+1)