It is known that the sum of the first n terms of the sequence an is Sn, and an + 2Sn = 4N (n ∈ n +). 1) to find the general term formula 2 of the sequence an It is known that the sum of the first n terms of an is Sn, and an + 2Sn = 4N (n ∈ n +) 1) The general term formula of the sequence an 2) If BN = Nan, find the first n terms of BN and TN

It is known that the sum of the first n terms of the sequence an is Sn, and an + 2Sn = 4N (n ∈ n +). 1) to find the general term formula 2 of the sequence an It is known that the sum of the first n terms of an is Sn, and an + 2Sn = 4N (n ∈ n +) 1) The general term formula of the sequence an 2) If BN = Nan, find the first n terms of BN and TN

When n = 1, a1 + 2S1 = 3A1 = 4 × 1 = 4A1 = 4 / 3N ≥ 2, an + 2Sn = 4nsn = (- an + 4N) / 2 = - an / 2 + 2nan = SN-S (n-1) = - an / 2 + 2n + a (n-1) / 2 - 2 (n-1) 3an = a (n-1) + 43an-6 = a (n-1) - 2 (An-2) / [a (n-1) - 2] = 1 / 3, which is a fixed value

Let the sum of the first n terms of sequence a n be the square of Sn = n-4n + 1, and find the general term formula

Sn=n^2-4n+1
Sn-1=(n-1)^2-4(n-1)+1
Sn=Sn-1+an
Then an = sn-sn-1 = n ^ 2-4n + 1 - (n-1) ^ 2 + 4 (n-1) - 1 = 2n + 3

Let the number sequence an, Sn = - 4N square + 25N + 1 (1) to find the general formula (2) to find the value of an when SN is the largest. Detailed process, thank you!

Sn=-4n²+25n+1
an=Sn-S(n-1)=-4n²+25n+1-[-4(n-1)²+25(n-1)+1]=29-8n
Sn=-4(n-25/8)²+641/16
n=3
S (3) = 40 is the maximum
a3=29-24=5