Question

已知等差数列｛an｝的前n项和为Sn＝n2＋pn＋q（p,q∈R）,且a2,a3,a5成等比数列

Answer 1

(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成等比数列

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)