The sum of the first n terms of the arithmetic sequence {an} is Sn = 4n2-25n

The sum of the first n terms of the arithmetic sequence {an} is Sn = 4n2-25n

When n ≥ 2, an = sn-sn-1 = 4 (2n-1) - 25 = 8n-29
When n ≤ 3, that is, when an < 0
Tn=-Sn=25n-4n²
When n ≥ 4, that is, when an > 0
Tn=Sn-2S3=4n²-25n+78

Let {an} be the arithmetic sequence, the sum of the first n terms is Sn, known as S7 = 7, S15 = 75, TN is the sum of the first n terms of the sequence {Sn / N}, find TN

Let the tolerance of {an} be DSN = n (a1 + an) / 2, because S7 = 7, S15 = 75, so S7 = 7 (a1 + A7) / 2 = 7a4 = 7, S15 = 15 (a1 + A15) / 2 = 15a8 = 75, so A4 = 1, A8 = 5, so d = (a8-a4) / 4 = (5-1) / 4 = 1, so A1 = a4-3d = 1-3 * 1 = - 2An = a1 + (n-1) d = - 2 + (n-1) * 1 = n-3, so Sn = n (a1 + an) / 2 = n (- 2 + n -

Let {an} be the arithmetic sequence, Sn be the sum of the first n terms, so as to know that S7 = 7, S15 = 75, tn be the sum of the first n terms of the sequence {Sn / N}, and find TN

Sn = n * a1 + n * (n-1) * D / 2, A1 is the first term, D is the tolerance
S7=7a1+21d=7
S15=15a1+105d=75
The solution is A1 = - 2, d = 1
Sn=-2n+n*(n-1)/2
Sn/n=-2+(n-1)/2
So Sn / N is an arithmetic sequence, the first term is - 2, and the tolerance is 1 / 2
So TN = - 2 * n + n * (n-1) * 1 / 2 / 2
=-2n+n(n-1)/4