In the sequence {an}, A1 = 2, an + 1 = an + ln (1 + 1n), then an = 1___ ．

In the sequence {an}, A1 = 2, an + 1 = an + ln (1 + 1n), then an = 1___ ．

A1 = 2 + ln1, A2 = 2 + LN2, A3 = 2 + LN2 + ln32 = 2 + Ln3, A4 = 2 + Ln3 + ln43 = 2 + ln4, thus conjecturing an = 2 + lnn. It is proved by mathematical induction that: ① when n = 1, A1 = 2 + ln1, holds. ② suppose that the equation holds when n = k, that is AK = 2 + LNK, then when n = K + 1, AK + 1 = AK + ln (1 + 1K) = 2 + LNK + LNK +

Let f = log2x - logx2 (0 ＜ x ＜ 1), and the sequence "an" satisfy that f (the power of an of 2) = 2n, n belongs to the general term of positive integer to find an

F (2 ＾ an) = 2n ＾ log2 2 ＾ an-log2 ＾ an 2 = 2n ＾ an-1 / an = 2n multiply an on both sides to get: an-2nan-1 = 0, an = n ± √ (n + 1) ∵ 0 ＾ x ＾ 0 ＾ 2 ＾ an ＾ 1 ＾ an ＾ 0 ＾ an = n - √ (n-1)

Given the function f (x) = ax − 5 (x > 6) (4 − A2) x + 4 (x ≤ 6), the sequence {an} satisfies an = f (n), (n ∈ n +), and {an} is a monotone increasing sequence, then the value range of real number a is ()
A. （7，8）B. [7，8）C. （4，8）D. （1，8）

We know that A1 = 4 − A2 + 4 = 8 − A2, A2 = 12 − a, A6 = 28-3a, a7 = a7-5, ∵ {an} is a monotone increasing sequence, ∵ 8 − A2 ＜ 12 − aa7 − 5 ＞ 28 − 3a, the solution is 4 ＜ a ＜ 8