The arithmetic sequence an, S7 = 7. S15 = 75. TN is the sum of the first n terms of {Sn / N}, and TN

The arithmetic sequence an, S7 = 7. S15 = 75. TN is the sum of the first n terms of {Sn / N}, and TN

a1+a7=2
a1+a15=10
So: a15-a7 = 8D = 8
D = 1
a1+a7=a1+a1+6d=2
2a1+6=2
Result: A1 = - 2
Sn=n²/2-5n/2
Sn / N = n / 2-5 / 2, arithmetic sequence
S1 / 1 = - 2; tolerance is 1 / 2;
Therefore, TN = n & # / 4-9n / 4;

{an} is the arithmetic sequence S7 = 7 S15 = 75 TN is the sum of the first n of the sequence {Sn / N} to find TN

Let an = a1 + (n-1) d
In S7 = A1N + n (n-1) d / 2 = 7a1 + 21d = 7
Similarly, S15 = 15a1 + 105d = 75
The simultaneous solution is A1 = - 2, d = 1
So Sn = - 2n + n (n-1) / 2
So {Sn / N} = - 2 + (n-1) / 2
SO 2 {Sn / N} = - 4 + (n-1), that is, 2 times {Sn / N} is an arithmetic sequence with - 4 as the first term and 1 as the tolerance
So 2tn = - 4N + n (n-1) / 2
So TN = - 2n + n (n-1) / 4

Let {an} be an arithmetic sequence. SN is the sum of the first n terms of the sequence {an}. It is known that S7 = 7, S15 = 75. TN is the sum of the first n terms of the sequence {Sn / N}. Find TN

Because {an} is the sum of the first n terms of {an}, S7 = 7, S15 = 75, so S7 = 7 (a1 + A7) / 2 = 7a4 = 7s15 = 15 (a1 + A15) / 2 = 15a8 = 75, so A4 = 1, a8 = 5, so A8 = A4 + 4D = 1 + 4D = 5, so d = 1, so A1 = a4-3d = 1-3 = - 2, so an = a1 + (n-1) d = n-3, so Sn = n (a1 + an) / 2 = n (N-5) / 2

An = 2 * n + 1, BN = 1 / (an ^ 2-1), find the first n terms of sequence B and TN

A:
bn=1/(an^2-1)
=1/[(2n+1)^2-1]
=1/(4n^2+4n)
=1/[4n(n+1)]
=1/4[1/n-1/(1+n)]
So TN = B1 + B2 +... + BN
=1/4[1/1-1/2+1/2-1/3+...+1/n-1/(1+n)]
=1/4[1-1/(1+n)]
=n/(4+4n)