(1)
Sn = n^2+pn+q
n=1 , a1= p+q+1
an = Sn - S(n-1)
= 2n-1 + p (1)
a1= 1+p = p+q+1
=>q=0
from (1)
a2 = 3+p
a3 =5+p
a5 = 9+p
a2,a3 ,a5成等比數(shù)列
a2.a5 = (a3)^2
(3+p)(9+p)=(5+p)^2
12p+27 = 10p+25
p=-1
(2)
an = 2n-2
an+logn=logbn
2n-2 = log(bn/n)
bn/n = 2^(2n-2)
bn = n.2^(2n-2)
let
S = 1.2^0 + 2.2^2 + .+n.2^(2n-2)(2)
4S = 1.2^2 + 2.2^4 + .+n.2^(2n)(3)
(3)-(2)
3S = n.2^(2n) - [ 1 + 2^2+2^4+...+2^(2n-2) ]
= n.2^(2n) - (1/3)(2^(2n) -1)
S = 3n.2^(2n) - (2^(2n) -1)
= 1 + (3n-1).2^(2n)
Tn = b1+b2+...+bn=S =1 + (3n-1).2^(2n)