It is known that the product of the third, fifth, and seventh terms of the increasing equal ratio sequence {an} is 512, and the three terms are subtracted by 1, 3, and 9 to form the equal difference sequence. (1) find the first term and the common ratio of {an}; (2) let Sn = A12 + A22 + +An2, find SN

It is known that the product of the third, fifth, and seventh terms of the increasing equal ratio sequence {an} is 512, and the three terms are subtracted by 1, 3, and 9 to form the equal difference sequence. (1) find the first term and the common ratio of {an}; (2) let Sn = A12 + A22 + +An2, find SN

(1) According to the properties of the equal ratio sequence, we can get A3 · A5 · A7 = A53 = 512, and the solution is A5 = 8. Let the common ratio of the sequence {an} be q, then A3 = 8q2, a7 = 8q2, and we can get (8q2-1) + (8q2-9) = 2 (8-3) = 10, and the solution is Q2 = 2 or 12. ∵ {an} is an increasing sequence, and we can get Q ＞ 1, ∵ Q2 = 2, and q = 2. Therefore, A5 = a1q4 = 4A1 = 8, and the solution is A1 = 2; (2) the general formula of {an} is an = A1 · Q from (1) N-1 = 2 × (2) n − 1 = (2) n + 1, ｜ an2 = [(2) n + 1] 2 = 2n + 1, we can get that {an2} is an equal ratio sequence with 4 as the first term and the common ratio equal to 2 +an2=4(1−2n)1−2=2n+2-4．