Let the sum of the first n terms of the sequence {an} be Sn, and 2An = Sn + 2n + 1 (n ∈ n *) (I) find A1, A2, A3; (II) prove that the sequence {an + 2} is an equal ratio sequence; (III) find the sum of the first n terms of the sequence {n · an} and TN

Let the sum of the first n terms of the sequence {an} be Sn, and 2An = Sn + 2n + 1 (n ∈ n *) (I) find A1, A2, A3; (II) prove that the sequence {an + 2} is an equal ratio sequence; (III) find the sum of the first n terms of the sequence {n · an} and TN

When n = 1, we get 2A1 = a1 + 3 and A1 = 3. When n = 2, we get 2A2 = (a1 + A2) + 5 and A2 = 8. When n = 3, we get 2A3 = (a1 + A2 + a3) + 7 and A3 = 18 (3 points) (II) prove: because 2An = Sn + 2n + 1, so 2An + 1 = Sn + 1 + 2n + 3 holds. By subtracting the two formulas, 2An + 1-2an = an + 1 + 2. So an + 1 = 2An + 2 (n ∈ n *), that is, an + 1 + 2 = 2 (an + 2) So the sequence {an + 2} is an equal ratio sequence with a1 + 2 = 5 as the first term and 2 as the common ratio (7 points) (III) from (II) & nbsp; we get: an + 2 = 5 × 2N-1, that is, an = 5 × 2n-1-2 (n ∈ n *). Then Nan = 5N · 2n-1-2n (n ∈ n *) Let the sum of the first n terms of the sequence {5N · 2N-1} be PN, then PN = 5 × 1 × 20 + 5 × 2 × 21 + 5 × 3 × 22 + +So 2pn = 5 × 1 × 21 + 5 × 2 × 22 + 5 × 3 × 23 + +So - PN = 5 (1 + 21 + 22 +...) +In other words, PN = (5n-5) · 2n + 5 (n ∈ n *) (11 points) so the first n terms of sequence {n · an} and TN = (5n-5) · 2n + 5-2 × n (n + 1) 2 are sorted out, TN = (5n-5) · 2n-n2-n + 5 (n ∈ n *) (13 points)