Given that the sum of the first n terms in the sequence an is Sn, satisfying Sn = 2an-1 and BN = 1-log1-2an, the general term formula of the sequence (an) and (BN) can be obtained Let the sum of n terms of sequence (anbn) be TN, and find the intersection of TN

RELATED INFORMATIONS

• 1. Let the sum of the first n terms of the sequence an be Sn, and S1 = 2, Sn + 1-sn = Sn + 2 = BN (n ∈ n *) 1 be positive: the sequence BN is an equal ratio sequence; Let the sum of the first n terms of the sequence an be Sn, and S1 = 2, Sn + 1-sn = Sn + 2 = BN (n ∈ n *) 1 be positive: the sequence BN is an equal ratio sequence The second question is to find the general term formula of sequence an To the left of the equal sign is small (n + 1)
• 2. Given sequence {an}, where a1 = 1, an = 3n-1 · an-1 (n ≥ 2, n ∈ n), the first n terms of sequence {BN} and Sn = log3 (an9n) where n ∈ n *. (1) find the general term formula of sequence {an}; (2) find the general term formula of sequence {BN}; (3) find TN = | B1 | + | B2 | + +|bn|．
• 3. If the sequence {an} is an arithmetic sequence and the first term A1 > 0, a2005 + a2006 > 0, the maximum natural number n of a2005a20060 is
• 4. In sequence an, A1 = 2, an = 1 / 3A (n-1) - 2 (n is greater than or equal to 2), then a2013 is
• 5. In the sequence {an} (subscript, the same after), A1 = 1, A2 = 2, the sequence {Anan + 1} is an equal ratio sequence with common ratio Q (Q > 0) Problem: find the first 2n terms and s2n of sequence {an}
• 6. The sum of the first n terms of the arithmetic sequence {an} is SN. If S15 is a definite constant, the following expressions are also definite constants () A. a2+a13B. a2a13C. a1+a8+a15D. a1a8a15
• 7. Let Sn be the sum of the first n terms of the sequence an, Sn = kn * 2 + N, n ∈ n *, where k is a constant, then find A1, an
• 8. Given the sequence {an}, where a1 = 10, and when n ≥ 2, an = 5an-1 / 6an-1 + 5, find the general term formula of the sequence {an} Given the sequence {an}, where a1 = 10, and when n ≥ 2, an = (5an-1) / (6an-1) + 5, find the common term of the sequence {an}
• 9. In the sequence {an}, if A1 = 2 / 3, then a (n + 1) = 1 / (n + 1) (n + 2) is equal to a(n+1)=1/(n+1)(n+2)+an
• 10. It is known that a, B, X1 and X2 are mutually unequal positive numbers, Y1 = (ax1 + bx2) / (a + b), Y2 (BX1 + AX2) / (a + b), try to analyze the size of y1y2 and x1x2, and say
• 11. Let {BN} and {an} satisfy {BN} = 3an, n ∈ n *, and A1 = 1. (1) judge what kind of sequence {an} is and give the proof; (2) if CN = 1anan + 1, find the sum of the first n terms of {CN}
• 12. It is known that the real number sequence an is equal ratio sequence and the common ratio is Q It is known that the real number sequence a (n) is an equal ratio sequence and the common ratio is Q. for all positive integers n > 1, there is: ((a (n + 1)) (s (n-1)) + (a (n-1)) (s (n + 1))) / 2
• 13. 1. Given the sequence {an} A1 = - 2, an + 1 = (1 + an) / (1-an), then a2011= A.-2 B.-1/3 C.-0.5 D.0.5
• 14. Given the sequence {an}, A1 = 1, a (n + 1) = an + 2 / an + 1. Prove that when n is greater than or equal to 2 and N belongs to positive integer, there is 1
• 15. In the sequence, A1 = 8, A4 = 2, and satisfy an + 2-2an + 1 + an = 0. It is proved that {an} is an arithmetic sequence
• 16. Let the sum of the first n terms of the sequence {an} be Sn, and (3-m) Sn + 2man = m + 3 (n ∈ n *). Where m is a constant, m ≠ - 3 and m ≠ 0, please solve one step (2) If the common ratio of sequence {an} satisfies q = f (m) and B1 = A1, BN = 32 F (bn-1) (n ∈ n *, n ≥ 2) bn }For the arithmetic sequence, and BN If B 1 = a 1 = 1, q = f (m) = 2m m + 3, n ∈ N and N ≥ 2, BN = 32 f (bn-1) = 32 & # 8226; 2bn-1 bn-1 + 3, can we explain it? (# 8658); we get bnbn-1 + 3bn = 3bn-1 & # 8658; 1bn-1 bn-1 = 13.; {1bn} is the first term of 1, and 13 is the arithmetic sequence of tolerance, so BN = 3N + 2 Proving {1 / BN} as arithmetic sequence And find BN
• 17. It is known that {an} is an equal ratio sequence with the first term A1, the common ratio Q, (q is not equal to 1, greater than 0), the sum of the first n terms is Sn, 5 * S2 = 4 * S4, let BN = q + Sn (1) It is known that {an} is an equal ratio sequence whose first term is A1, common ratio is Q, (q is not equal to 1, greater than 0). The sum of the first n terms is Sn, and 5 * S2 = 4 * S4. Let BN = q + Sn (1) find out whether the Q (2) sequence BN can be an equal ratio sequence. If we can find out Q, we can't give the reason,
• 18. 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}
• 19. Given the first n terms of {an} and Sn = an2 + BN (a is not equal to 0), find the general term formula of {an} and prove that the sequence is arithmetic sequence
• 20. Let the sum of the first n terms of the arithmetic sequence {an} be Sn, if S9 = 72, then A2 + A4 + A9 = () A. 12B. 18C. 24D. 36