If Sn / TN = (2n + 3) / (3n-1), calculate A9 / B9

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• 1. The arithmetic sequence an, S7 = 7. S15 = 75. TN is the sum of the first n terms of {Sn / N}, and TN
• 2. The sequence an = 6n-3, BN = 5n-4, if an is less than or equal to 1000, BN is less than or equal to 1000, a new sequence cn is composed of both the terms in an and BN. How many terms does an have in total, and what is the sum of all these terms
• 3. The known sequence {an} satisfies the following conditions: A1 = 1, a (n + 1) = 2An + 1, n ∈ n * (3) proof: n / 2-1 / 3
• 4. Let an = n ^ 2 + λ n, A1
• 5. In the equal ratio sequence {an}, a1 + A7 = 65, A3 · A5 = 64, and an + 1 & lt; an (1) finds the general term of the sequence {an}. (2) finds the first five terms and s of the sequence {an}
• 6. The known sequence {an} satisfies A1 = 1; an = a1 + 2A2 + 3a3 +... + (n-1) a (n-1); (n > = 2); find the general term formula
• 7. The known sequence {an} satisfies A1 = 1, an = a1 + 1 / 2A2 + 1 / 3a3 +1 / (n-1) a (n-1), (n > 1, n ∈ n) a2=a1=1 n> When = 3 an+1=a1+1/2a2+.+1/n-1an-1+1/nan The two formulas are subtracted to obtain an + 1-an = 1 / Nan That is an + 1 = n + 1 / Nan That is, an + 1 / an = n + 1 / n an+1/a2=(an+1/an)(an/an-1).(a3/a2) =(n+1/n)(n/n-1).(3/2) =n+1/2 That is, an + 1 = n + 1 / 2 That is, an = n / 2 After obtaining an + 1 / an = n + 1 / N, why can't we directly use the formula of equal ratio sequence to find an
• 8. The known sequence {an} satisfies A1 = 1, an = a1 + 2A2 + 3a3 + +(n-1) an-1, then when n ≥ 2, the general term an = () A. n!2B. (n+1)!2C. n!D. （n+1）!
• 9. It is known that the sum of the first n terms of the sequence {an} is Sn, and a1 + 2A2 + 3a3
• 10. It is known that an = 2A (n-1) + 2 ^ n-1 (n ≥ 2) A1 = 5, A2 = 13, A3 = 33, A4 = 81. {(an + α) / 2 ^ n} is an arithmetic sequence, and a is obtained
• 11. It is known that the sum of the first n terms of two arithmetic sequences {an}, {BN} is Sn and TN respectively. If Sn / TN = (5N + 3) / (2n-1), then A9 / B9 is obtained Such as the title
• 12. Given the sum of the first n terms of the sequence {an} and Sn = - n ^ 2 + 4N 1) find the general term formula of an; 2) find the sum of the first n terms of the sequence {9-2an / 2 ^ n} Because the original problem is the quadratic of negative n, not the quadratic of negative n, so ~ A1 seems to be 3, not 5! The general formula is also wrong!
• 13. The first n terms of known sequence {an} and Sn = 1-5 + 9-13 + 17-21 +... + (- 1) ＾ n * (4n-3) Find S15 + s22-s31
• 14. Sequence {an} satisfies A1 = 1 / 2, a1 + A2 +.. + an = n square an, find an
• 15. Final warm-up (2) 17 questions, the first N-term sum of sequence {an} is Sn, known A1 = 1. An + 1 = (n + 2) Sn / N (n ∈ n *) It is proved that: (1) the equal ratio sequence of sequence {Sn / N}. (2) Sn + 1 = 4An focuses on the second question,
• 16. Given that the general term formula of sequence is an = 2n-47, then when Sn takes the minimum value, n=______ ．
• 17. Let the first n terms and Sn = (- 1) ^ n (2n ^ 2 + 4N + 1) - 1 of sequence {an} 1, find the general term formula an of sequence {an} 2, record BN = (- 1) ^ n / an, find the first n terms and TN of the sequence {BN}
• 18. Given that the first n terms of an and Sn are equal to the square of n minus n plus 1, the general term formula of an is obtained
• 19. Given that the sum of the first n terms of a sequence an is Sn, and an = 3 ^ n + 2n, then SN is equal to
• 20. If Sn = 2N-1, then A8=______ ．