The first n terms and Sn of the equal ratio sequence {an} belong to N + for any n, and the point (n, Sn) is in the function y = B ^ x + R (b > 0, and B is not equal to 1, R constant) (1) Finding the value of R (2) When B = 2, denote BN = n + 1 / 4An (n belongs to N +) and find the first n terms and TN of the sequence {BN}

The first n terms and Sn of the equal ratio sequence {an} belong to N + for any n, and the point (n, Sn) is in the function y = B ^ x + R (b > 0, and B is not equal to 1, R constant) (1) Finding the value of R (2) When B = 2, denote BN = n + 1 / 4An (n belongs to N +) and find the first n terms and TN of the sequence {BN}

(1) Sn = A1 (1-Q ^ n) / (1-Q) and point (n, Sn) is on y = B ^ x + R, so Sn = B ^ n + R gets A1 / 1-Q = 1, B = - Q r = 1 (2) B = 2, so sequence {an} is common ratio q = - 2, first term A1 = - 1, so general term an = - (- 2) ^ n-1, then the first N phase and TN of general term BN = n + 1 / 4An sequence BN can be regarded as the first n of arithmetic sequence n

Let an, an + K (k is a constant) be equal ratio sequence, if A1 = 2, Sn is the sum of the first n terms of an, and K is not equal to 0, then s (3n-1) - BN=

Let the common ratio of an equal ratio sequence be Q
An + K (k is a constant) is an equal ratio sequence
So:
The common ratio of BN = an + K (k is a constant) is: (2q ^ n + k) / (2q ^ (n-1) + k)
(2q ^ n + k) / (2q ^ n-1) + k) is a constant not equal to 0
Namely:
(2q^n+k)/(2q^(n-1) +k)
=[q( 2q^(n-1) +k ) +k-kq ] /(2q^(n-1) +k)
=Q + K (1-Q) / (2q ^ (n-1) + k) is a constant not equal to 0
That is, the value of K (1-Q) / (2q ^ (n-1) + k) has nothing to do with the value of n
Let n = 1 K (1-Q) / (2q ^ (n-1) + k) = 1-Q (k is not equal to 0)
Let n = 2K (1-Q) / (2q ^ (n-1) + k) = K (1-Q) / (2q + k)
Let 1-Q = K (1-Q) / (2q + k) (q is not equal to 0)
Q = 1
Substituting q = 1 into the formula k (1-Q) / (2q ^ (n-1) + k) = 0, its value is independent of the value of N and satisfies the condition
That is: the common ratio of BN is (2q ^ n + k) / (2q ^ (n-1) + k)
=q+ k(1-q)/ (2q^(n-1) +k)
=q
=1
That is BN = a1 + k = 2 + K
S(3n-1)-bn=2（3n-1）--(2+k)=6n-k

The first n terms and Sn of sequence [an] are equal to 2 * n-1, the sequence [BN] satisfies: B1 = 3, BN + 1 = an + BN, n belongs to n *. 1. It is proved that sequence [an] is an equal ratio sequence
2. Find the first n terms and TN of sequence [BN]

When 1, ∵ Sn = 2 & # 8319; - 1 ∵ A & # 8321; = S & # 8321; = 1sn-1 = 2 * (n-1) - 1 ∵ an = sn-sn-1 = 2 & # 8319; - 2 * (n-1) = 2 * (n-1) n = 1, a & # 8321; = 1, in accordance with ∵ an = 2 * (n-1), it is equal ratio sequence 2, BN + 1 = an + bnbn BN BN = an-1 + bn-1 B &; = A &; + B &; left and right