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Given sequence {an}, where a1 = 1, an = 3n-1 · an-1 (n ≥ 2, n ∈ n), the first n terms of sequence {BN} and Sn = log3 (an9n) where n ∈ n *. (1) find the general term formula of sequence {an}; (2) find the general term formula of sequence {BN}; (3) find TN = | B1 | + | B2 | + +|bn|．
(1) Because an = 3n-1 · an-1 (n ≥ 2, n ∈ n), log3an = log3an-1 + (n-1), an = 3n-1 · an-1 (n ≥ 2, n ∈ n), log3an-log3a1 = 1 + 2 + 3 + +(n-1) = n (n-1) 2, then an = 3N (n-1) 2 (2) and B1 = S1 = - 2, when n ≥ 2, BN = sn-sn-1 = n-3, n = 1 is also suitable, so the general formula of sequence {BN} is BN = n-3 (n ∈ n *) (3) when BN = n-3 ≤ 0, that is, n ≤ 3, TN = - Sn = 5n-n22, when BN = n-3 > 0, that is, n > 3, TN = | B1 | + | B2 | + +|bn|=(b1+b2+… +BN) - (B1 + B2 + B3) = sn-2s3 = n2-5n + 122. To sum up, TN = 5n-n22 (n ≤ 3, and N ∈ n *) n2-5n + 122 (n > 3, and N ∈ n)
Given sequence {an}, where a1 = 1, a (n + 1) = 3 ^ (2n-1) * an (n ∈ n), the first n terms of sequence {BN} and Sn = log3 (an / 9 ^ n) (n ∈ n)
Find an BN
A1 = 1A2 = 3 ^ 1 * a1a3 = 3 ^ 3 * A2... An = 3 ^ (2n-3) * A3A (n + 1) = 3 ^ (2n-1) * an two sides are multiplied to get a (n + 1) = 3 ^ [1 + 3 + 5 +... + (2n-1)] = 3 ^ (n ^ 2) an = 3 ^ [(n-1) ^ 2] Sn = log3 (an / 9 ^ n) = log3 (an) - log3 [3 ^ (2n)] = (n-1) ^ 2-2nn = 1. B1 = S1 = - 2, n > 1, BN = SN-S (n-1) = 2n
It is known that the sum of the first n terms of the sequence {an} is Sn, and A1 = 1, a (n + 1) = Sn / 2,
It is known that the sum of the first n terms of the sequence {an} is Sn, and A1 = 1, a (n + 1) = Sn / 2
When BN = log3 / 2 (3a (n + 1)), prove the first n terms and TN = n / 1 + n of sequence {1 / BNB (n + 1)}
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