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Look at the ratio of the (n+1)-st term to the n-th. After cancellation, you have [(2^n + 1)*(2^n + 3)*...*(2^(n+1) - 1)]/[2*(n + 1)] which is > 1 when n >= 2. Therefore, a_(n+1) > a_n, so the sequence is eventually increasing.

2007-09-19 01:11:06 · answer #1 · answered by Tony 7 · 0 0

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