Given that the sum of the first n terms of the sequence {an} is Sn, and S (n + 1) = 4An + 2, A1 = 1, let CN = an / 2 ^ n, prove that the sequence {CN} is an arithmetic sequence
S (n + 1) = 4An + 2Sn = 4A (n-1) + 2A (n + 1) = s (n + 1) - Sn = 4 (an-a (n-1)) a (n + 1) - 2An = 2 [an-2a (n-1)] [a (n + 1) - 2An] / [an-2a (n-1)] = 2 {an-2a (n-1)} is the arithmetic sequence A1 = 1S2 = 4 + 2 = 6, A2 = s2-a1 = 6-1 = 5an-2a (n-1) = [a2-2a1] * 2 ^ (n-2) = 3 * 2 ^ (n-2) a
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