If the first n terms of the sequence {an} and Sn = 4n2-n + 2, then the general term formula of the sequence is an=______ ．

∵ an = sn-sn-1 (n ≥ 2), ∵ an = 8n-5 (n ≥ 2), when n = 1, A1 = S1 = 5, ∵ when n = 1, an = 5; when n ≥ 2, an = 8n-5. So the answer is: when n = 1, an = 1; when n ≥ 2, an = 8n-5

The general formula of sequence {an} is an = 1 / (4n-3) (4N + 1) for Sn

An=1/(4n-3)(4n+1)=[1/(4n-3)-1/(4n+1)]/4
Sn=[1-1/5+1/5-1/9+...+1/(4n-7)-1/(4n-3)+1/(4n-3)-1/(4n+1)]/4
=[1-1/(4n+1)]/4
=n/(4n+1)

It is known that the sum of the first n terms of the sequence an is Sn, and an + 2Sn = 4N (n ∈ n +). Find the general term formula of the sequence an, (2) if BN = Nan, find the sum of the first n terms of the sequence BN and TN

”On the formula of finding general term by term and sum
When an + 2Sn = 4N, n = 1, the solution is A1 = 4 / 3
an-1+2Sn-1=4(n-1)
By subtracting an: n = 2 A-1 +
3an = (an-1) + 4 (constructing equal ratio sequence method)
3 (An-2) = (an-1) - 2 (An-2 is the nth minus 2, an-1 is the nth)
{An-2} is an equal ratio sequence with - 2 / 3 as the first term and 1 / 3 as the common ratio
That is to say, An-2 = (- 2 / 3) · (1 / 3) ^ (n-1), an = (- 2 / 3) · (1 / 3) ^ (n-1) + 2
(2) From the characteristics of (1) and BN expansion, we know that the method of summation is grouping summation
The first part is equal difference multiplication equal ratio (dislocation subtraction), and the second part is the sum of the first n terms of the sequence {2n}
I don't know how to ask

If the first n terms of the sequence {an} and Sn = 4n2-n + 2, then the general term formula of the sequence is an=______ ．