Given the arithmetic sequence {an}, BN = 1 / Sn and a3b3 = 1 / 2, S5 + S3 = 21, find BN

Given the arithmetic sequence {an}, BN = 1 / Sn and a3b3 = 1 / 2, S5 + S3 = 21, find BN

Let AK = a1 + (k-1) d (let d be the tolerance) A3 = a1 + 2D, Sn = n * a1 + n (n-1) / 2
So S3 = 3A1 + 3D, because BN = 1 / Sn and a3b3 = 1 / 2, we deduce a = D, so Sn = D * n (n + 1) / 2
Because S5 + S3 = 21, d = 1,
So Sn = D * n (n + 1) / 2
So BN = 2 / [n (n + 1)]

Let {an} be an equal ratio sequence with the first term of 1. If {1 / [2An + a (n + 1)]} is an arithmetic sequence, then [1 / (2A1) + 1 / (A2)] + [1 / (2A2) + 1 / (A3)] + [1 / (2A3) + 1 / (A4)] + +The value of [1 / (2a2012) + 1 / (a2013)] is equal to_______

Let A1 = 1, A2 = q, A3 = q square by equal ratio
1 / (2 + Q) + 1 / (2q + Q) = 2 / (2q + Q) = 1
That is, the common ratio is 1 and the tolerance is 0
Original formula = 2012 / 3

Write a general term formula so that the first four terms of the sequence are:
[1+(1/2^2)]
[1-(3/4^2)]
[1+(5/6^2)]
[1-(7/8^2)]
To avoid the waste of points, please attach the expected additional points to the answers

Take a look
An=1-[(-1)^n]*[(2n-1)/(2n)^2] n=1,2,...
If you think it's right, just give it to me