If {an} is an equal ratio sequence, A2 = 2, A5 = 14, then A1A2 + a2a3 + a3a4 + +an-1an=______ ．

If {an} is an equal ratio sequence, A2 = 2, A5 = 14, then A1A2 + a2a3 + a3a4 + +an-1an=______ ．

∵ {an} is an equal ratio sequence, A2 = 2, A5 = 14, let its common ratio be q, then Q3 = A5, A2 = 18, q = 12, let BN = an · an + 1, bnbn − 1 = an + 1An − 1 = Q2 = 14 (n ≥ 2) and A1 = a2q = 4, then ∵ {BN} is an equal ratio sequence with the first term of 8 and the common ratio of 14, let the sum of the first n terms be Sn, then sn-1 = A1A2 + a2a3 + a3a4 + +An-1an = 323 (1 − 14N − 1), so the answer is: 323 (1 − 14N − 1)

Given that {an} is an equal ratio sequence, A2 = 2, A5 = 14, then A1A2 + a2a3 + +anan+1=（　　）
A. 16（1-4-n）B. 16（1-2-n）C. 323（1-4-n）D. 323（1-2-n）

From A5 = 14 = A2 · Q3 = 2 · Q3, the solution is q = 12. The sequence {Anan + 1} is still an equal ratio sequence: the first term is A1A2 = 8, the common ratio is 14, so A1A2 + a2a3 + +Anan + 1 = 8 [1 - (14) n] 1-14 = 323 (1-4-n), so C

Given that {an} is an equal ratio sequence, A2 = 2, A5 = 14, then A1A2 + a2a3 + +anan+1=（　　）
A. 16（1-4-n）B. 16（1-2-n）C. 323（1-4-n）D. 323（1-2-n）

From A5 = 14 = A2 · Q3 = 2 · Q3, the solution is q = 12. The sequence {Anan + 1} is still an equal ratio sequence: the first term is A1A2 = 8, the common ratio is 14, so A1A2 + a2a3 + +Anan + 1 = 8 [1 - (14) n] 1-14 = 323 (1-4-n), so C

Given that an is an equal ratio sequence, A2 = 2, A5 = 1 / 4, then A1A2 + a2a3 + +Ana (n + 1) = why ana (n + 1) / a (n-1) * an = q ^ 2?

Brother, you should take a look at the definition and formula of the sequence of equal ratio, A5 / A2 = q ^ 3 = 1 / 8, so q = 1 / 2 (from A2 = 2, q = 1 / 2, A1 = 4) let BN = ana (n + 1) BN = ana (n + 1) = a1q ^ (n-1) * a1q ^ n = (1 / 2) ^ (2n-5) B (n-1) = ana (n-1) = (1 / 2) ^ (2n-7) ana (n + 1) / ana (n-1) = (1 / 2) ^ 2 = q ^ 2, let t