In the sequence {an}, Sn = 9n-n ^ 2, find the absolute value of TN = a1 + the absolute value of A2 +. + the absolute value of an

a1=S1=9-1=8.
an =Sn-S(n-1)= 9n-n^2-[9(n-1)-(n-1)^2]
=10-2n.
∴an=10-2n.(n∈N*)
Therefore, the first four terms of the sequence are positive, the fifth is 0, and the subsequent terms are negative
When n ≤ 5, TN = | A1 | + | A2 | + +|an|=Sn=9n-n^2.
When n > 5, TN = S5 + [- (sn-s5)] = 2 S5 SN
=2×5(8+0)/2-(9n-n^2)= n^2-9n+40.

RELATED INFORMATIONS

1. In the arithmetic sequence {an}, Sn is the sum of the first n terms, S10 = S8, A1 = 17 / 2. Find out the sum of the absolute values of A1 ~ an
2. Given the arithmetic sequence an, a (n + 2) = 2A (n + 1) - 3N + 1, what is A5 equal to
3. In the arithmetic sequence [an], if an is equal to one of the integrals of (3n-2) (3N + 1), then what are the first n terms and Sn of [an]?
4. If the sum of the first n terms of the arithmetic sequence {an}, {BN} is an, BN, and an / BN = n / 3n-2, then A5 / B5 =?
5. In the arithmetic sequence {an}, A1 = - 60, A17 = - 12, find the sum of the first n terms of the sequence {an} From A1 = - 69, A17 = - 12, we can get the tolerance d = 3, an = 3n-63, which can be divided into two cases, n ＞ 20 and N ≤ 20. Let Sn and s'n represent the sum of the first n terms of {an} and {an} respectively. When n ≤ 20, [I will do it], but when n ＞ 20, s'n = - S20 + (sn-s20) = sn-2s20,
6. In the arithmetic sequence {an}, A1 = 60. A17 = 12, find the sum of the first n terms of the sequence {an}
7. In the known sequence {an}, A1 = 1, an + 1 = 2An + 1, let BN = an + 1-an. (1) prove that the sequence {BN} is an equal ratio sequence; (2) let the sum of the first n terms of the sequence {Nan} be Sn, and find the minimum value of positive integer n that makes Sn + n (n + 1) 2 ＞ 120
8. In the equal ratio sequence an, a1 + a3 = 10, A4 + A6 = 5 / 4, the value of common ratio q is
9. Given that the sequence {an} is an equal ratio sequence, a1 + a3 = 10, A4 + A6 = 5 / 4, find the value of A5 and the first n terms and S6
10. In the equal ratio sequence {an}, A1 = 2, A4 = 16 are known
11. It is known that the sum of the first n terms of the arithmetic sequence is Sn, and S13 < 0, S12 > 0, then the term with the smallest absolute value in the sequence is () A. Item 5 B. item 6 C. item 7 d. item 8
12. In the arithmetic sequence {an}, A2 + A8 = 16, a3a7 = 48, find the general term formula of the sequence, and explain when D
13. In the known arithmetic sequence {an}, a3a7 = - 16, A4 + A6 = 0, ① find {an} ② if D
14. In the known arithmetic sequence {an}, D ＞ 0, a3a7 = - 16, A2 + A8 = 0, let TN = | A1 | + | A2 | + +|An |. Find the general formula an of (I) {an}; (II) find TN
15. In the arithmetic sequence {an}, A1 = - 60, A17 = - 12. (1) find the general term an, (2) find the sum of the first 30 terms of the sequence
16. (1) Let the general term formula of arithmetic sequence {an} be 3n-2, and find the sum formula of its first n terms
17. Let the general term formula of arithmetic sequence {an} be 3n-2, and find the sum formula of its first n terms
18. If the general term formula of sequence an is known as an = 4N + 1, then the first n terms and Sn are equal to
19. The general term formula of sequence {an} is an = (- 1) ^ (n-1) * (4n-3), and the sum of the first n terms of sequence {an} can be obtained Solution: when n is even, Sn = (1-5) + (9-13) + (17-21) + ·· + [(- 1) ^ (n-2) (4n-7) + (- 1) ^ (n-1) (4n-3)] = - 4 * (n / 2) = - 2n When n is an odd number Why do you do this? I don't understand. Can you explain it? Thank you I still don't understand
20. Let the sum of the first n terms be Sn, then S100=