The sum of the first n terms of the proportional sequence {an} is Sn if s2n = 3 (a1 + a3 +...) +A2n-1), a1a2a3 = 8, then A10 equals () A. -512B. 1024C. -1024D. 512

The sum of the first n terms of the proportional sequence {an} is Sn if s2n = 3 (a1 + a3 +...) +A2n-1), a1a2a3 = 8, then A10 equals () A. -512B. 1024C. -1024D. 512

By using the properties of the equal ratio sequence, we can get that a1a2a3 = A23 = 8 & nbsp; that is, A2 = 2 because s2n = 3 (a1 + a3 +...) +So when n = 1, S2 = a1 + A2 = 3A1, then A1 = 1, q = 2, so A10 = 1 × 29 = 512, so D

It is known that the sum of the first n terms of the arithmetic sequence (an) is Sn = n ^ 2 + PN + Q (P, Q ∈ R), and A2, A3 and A5 are proportional to the sequence (1). Find the value of P and Q (2). If the sequence BN satisfies an + log2n = log2bn, find the first n terms and TN of the sequence BN

(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 are equal ratio sequence
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)

Let the first N-term product of sequence {an} be TN, TN = 1-an, and CN = 1 / TN (1) prove that sequence {CN} is an arithmetic sequence
(2) Is there any positive integer m, n (1 ＜ m ＜ n) in the general term formula (3) of the sequence {an}, so that A1, am, an are equal proportion sequence? If it exists, find out all the values of M, N. if it does not exist, say the reason. Just do the third question. I deduce that the formula can't solve m, n

T1 = A1 = 1-a1 2A1 = 1 A1 = 1 / 2a1a2... An = TN = 1-an (1) A1A2... A (n-1) = TN-1 = 1-A (n-1) (2) (1) / (2) an = (1-an) / [1-A (n-1)], we get [2-A (n-1)] an = 1An = 1 / [2-A (n-1)] A2 = 1 / (2-1 / 2) = 2 / 3. Suppose that when n = K (K ∈ n +, and K ≥ 2), AK = K / (K + 1), then when n = K + 1

It is known that the sum of the first n terms of sequence {an} is TN, and satisfies TN = 1-an, the sum of the first n terms of sequence {BN} is Sn, Sn = 1-bn, let CN = 1 / TN, and prove that {CN} is an arithmetic sequence
It is proved that {CN} is the general formula for finding {an} in arithmetic sequence 2. If TN (NbN = n-2) ≤ kn holds for n ∈ n *, the value range of real number k can be found

T (n + 1) - TN = a (n + 1) = 1-A (n + 1) - 1 + an, that is, a (n + 1) = an / 2. T1 = 1-a1, A1 = 1 / 2. An is an equal ratio sequence with the first term of 1 / 2 and the common ratio of 1 / 2, and an = (1 / 2) & # 8319;, similarly, BN = (1 / 2) & # 8319;. TN = 1 - (1 / 2) & # 8319;, CN = 1 / TN = 1 + 1 / (2 & # 8319; - 1). C (