Given that the sum of the first n terms of the sequence {an} is Sn, and Sn = n-5an-85, n ∈ n *, it is proved that {an-1} is an equal ratio sequence

∵ Sn = n-5an-85 ∵ an = SN-S (n-1) = n-5an-85 - (n-1) + 5A (n-1) + 85 = 1-5an + 5A (n-1) ∵ 6An = 5A (n-1) + 16an-6 = 5A (n-1) + 1-66 (an-1) = 5 (a (n-1) - 1) ∵ an-1 = 5 / 6 * (a (n-1) - 1) = 5 / 6 * B (n-1) ∵ {an-1} is an equal ratio sequence with a common ratio of 5 / 6,

RELATED INFORMATIONS

1. It is known that the sum of the first n terms of the sequence {an} is Sn, and Sn = n-5an-85, n belongs to n *, it is proved that {an-1} is an equal ratio sequence
2. It is known that the sum of the first n terms of the sequence {an} is Sn, satisfying an + Sn = 2n. (I) prove that the sequence {An-2} is an equal ratio sequence, and find out an; (II) let BN = (2-N) (An-2), find out the maximum term of {BN}
3. It is known that the sum of the first n terms of the sequence (an) is Sn, satisfying an + Sn = 2n. It is proved that the sequence (An-2) is an equal ratio sequence and an is obtained
4. Given the first n terms of sequence an and Sn = n (2n-1), it is proved that (an) is an equal ratio sequence
5. Let the first n terms of sequence {an} and Sn = 2an-2n (1) prove that sequence {an + 1-2an} is equal difference sequence (2) prove that sequence {an + 2} is equal ratio sequence (3) The general term formula of {an}
6. Let the first n terms of sequence {an} and Sn = 2an-2 ^ n (1) prove that {a (n + 1) - 2An} is the general term of finding {an} from the equal ratio sequence (2) The second question doesn't matter. Try to do it
7. In the equal ratio sequence, Sn is the sum of the first n terms, Sn = 2an-1, and an is obtained An = 2 (n-1) (n-1 power of 2)
8. If the sum of the first n terms of the sequence {an} is Sn = 2An + 3, then an is an equal ratio sequence Give reasons
9. Let the sum of the first n terms of the equal ratio sequence {an} be Sn, A1 = 2011, and an + 2An + 1 + an + 2 = 0 (n ∈ n ·), then s2012=______ ．
10. It is known that SN is the sum of the first n terms of the sequence an, and Sn = 2-2an 1. Prove that the sequence an is an equal ratio sequence 2. Find the general term formula of the sequence an 3. Find the first n terms and wn of the sequence {ANSN} We need a specific process
11. If the first n terms of the equal ratio sequence {an} and Sn = 3N + R, then r = () A. 0B. -1C. 1D. 3
12. The sequence {an} is an equal ratio sequence composed of real numbers, Sn = a1 + A2 + +An, then () A. Any term is not 0b. There must be one term that is not 0C. At most, there are finite terms that are 0d. Or none of them is 0, or there are infinite numbers that are 0
13. If. Sn = m * 3 ^ n + 1, then the real number M
14. For any real number n, Sn = 2 ^ n-1, then A1 ^ 2 + A2 ^ 2 + a3 ^ 2 +... + an ^ 2 is equal to?
15. If the first n terms of the equal ratio sequence {an} and Sn = T5 ^ (n-2) are known, then the value of the real number T is
16. In the sequence {an}, A1 is equal to 3, a (n + 1) is equal to a * a (n) + 1-A (a is not equal to 0), find a (n) and Sn
17. The sum of the first n terms of the sequence {(- 1) ^ n * a ^ 4N} (a is not equal to 0) is equal to
18. Given the first n terms of the sequence {an} and Sn = an − 1 (a ≠ 0), then the sequence {an} () A. It must be an arithmetic sequence B. it must be an arithmetic sequence C. It must be an arithmetic sequence or an arithmetic sequence D. It is neither an arithmetic sequence nor an arithmetic sequence
19. Sum: SN = 1 + 3x + 5x + 7x + +﹝2n-1﹞x^n-1
20. Sum by dislocation subtraction: sum: SN = 1 + 3x + 5x2 + 7X3 + +（2n-1）xn-1．