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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
The equal ratio sequence is an + 1 / an = q, which is a number independent of n
RELATED INFORMATIONS
- 1. 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）!
- 2. It is known that the sum of the first n terms of the sequence {an} is Sn, and a1 + 2A2 + 3a3
- 3. 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
- 4. In the sequence an, A1 = in, a (n + 1) = 2An + 3n-4, in is a real number. For any in, it is proved that an is not an equal ratio sequence BN = a (n + 1) - an + 3, try to judge whether BN is an equal ratio sequence The formula of finding the same term of an
- 5. Given that the sum of the first n terms of sequence {an} A1 = 2 is Sn and satisfies Sn sn-1 = 3an, the general term formula an of sequence {an} is obtained Given that the sum of the first n terms of the sequence {an} A1 = 2 is Sn and satisfies Sn + sn-1 = 3an, the general term formula an of the sequence {an} is obtained
- 6. Let f (x) satisfy 2F (x) - f (1 / x) = 4x-2 / x + 1, sequence {an} and {BN} satisfy A1 = 1, a (n + 1) - 2An = f (n), BN = a (n + 1) - an Finding the analytic expression of F (x) The general term formula of BN Is to compare the size of 2An and BN, and prove that
- 7. It is known that in the arithmetic sequence an, A4 = 4, the first 10 terms and S10 = 10, BN = (1 / C) are to the power of an, (C is a positive constant.) (1) Find an (2) It is proved that BN is an equal ratio sequence (3) Finding the first n terms and TN of sequence BN
- 8. In the arithmetic sequence {an}, a1 + a3 + A5 = 18, an-4 + An-2 + an = 108, Sn = 420, then n=______ ．
- 9. Given that the sum of the first n terms of the equal ratio sequence {an} is Sn = (2 ^ n) - 1, find the sum of the first n terms of the sequence {(an) ^ 2} and TN
- 10. 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
- 11. The known sequence {an} satisfies A1 = 1; an = a1 + 2A2 + 3a3 +... + (n-1) a (n-1); (n > = 2); find the general term formula
- 12. 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}
- 13. Let an = n ^ 2 + λ n, A1
- 14. The known sequence {an} satisfies the following conditions: A1 = 1, a (n + 1) = 2An + 1, n ∈ n * (3) proof: n / 2-1 / 3
- 15. 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
- 16. The arithmetic sequence an, S7 = 7. S15 = 75. TN is the sum of the first n terms of {Sn / N}, and TN
- 17. If Sn / TN = (2n + 3) / (3n-1), calculate A9 / B9
- 18. 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
- 19. 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!
- 20. The first n terms of known sequence {an} and Sn = 1-5 + 9-13 + 17-21 +... + (- 1) ＾ n * (4n-3) Find S15 + s22-s31