Let the sum of the first n terms of the sequence {an} be SN. Given A1 = 1, Sn + 1 = 4An + 2, find: (1) let BN = an + 1-2an, prove that the sequence {BN} is an equal ratio sequence 2) Finding the general term of an The second answer is (1) bn＝a(n＋1)－2an＝3•2^(n－1) ∴[a(n＋1)]/[2^(n＋1)]－(an)/(2^n)=3/4 The sequence {(an) / (2 ^ n)} is an arithmetic sequence with 1 / 2 first term and 3 / 4 tolerance ∴(an)/(2^n)＝1/2＋(n－1)3/4＝3/4n－1/4 That is, an = (3n-1) & _; 2 ^ (n-2) (n ∈ n *) Question: how to get [a (n + 1)] / [2 ^ (n + 1)] - (an) / (2 ^ n) = 3 / 4 and why

Let the sum of the first n terms of the sequence {an} be SN. Given A1 = 1, Sn + 1 = 4An + 2, find: (1) let BN = an + 1-2an, prove that the sequence {BN} is an equal ratio sequence 2) Finding the general term of an The second answer is (1) bn＝a(n＋1)－2an＝3•2^(n－1) ∴[a(n＋1)]/[2^(n＋1)]－(an)/(2^n)=3/4 The sequence {(an) / (2 ^ n)} is an arithmetic sequence with 1 / 2 first term and 3 / 4 tolerance ∴(an)/(2^n)＝1/2＋(n－1)3/4＝3/4n－1/4 That is, an = (3n-1) & _; 2 ^ (n-2) (n ∈ n *) Question: how to get [a (n + 1)] / [2 ^ (n + 1)] - (an) / (2 ^ n) = 3 / 4 and why

a(n＋1)－2an＝3•2^(n－1)
Divide both sides by 2 ^ (n + 1)
∴[a(n＋1)]/[2^(n＋1)]－(an)/(2^n)=3/4

If a1 + a3 = 10, A4 + A6 = 54, then the common ratio q is equal to ()
A. 2B. -2C. 12D. -12

According to the meaning of the question, let the common ratio be q, because a1 + a3 = 10, A4 + A6 = 54, so Q3 = A4 + A6, a1 + a3 = 18, q = 12, so choose C

Given the positive term sequence {an}, the first n terms and Sn satisfy 10sn = an2 + 5An + 6, and A1, A3, A15 are equal proportion sequence, find the general term an of the sequence {an}

∵ 10sn = an2 + 5An + 6, ① ∵ 10A1 = A12 + 5A1 + 6, the solution is A1 = 2 or A1 = 3. And 10sn-1 = an-12 + 5an-1 + 6 (n ≥ 2), ② from ① - ②, we get & nbsp; 10An = (an2-an-12) + 5 (an-an-1), that is, (an + an-1) (an-an-1-5) = 0 ∵ an + an-1 > 0, ∵ an-an-an-1 = 5 & nbsp; (n ∵