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Approximation of a sequence of real numbers

Source: Romanian IMO TST 2006, day 5, problem 3

May 23, 2006
floor functionlogarithmsfunctionalgebra proposedalgebra

Problem Statement

Let x1=1x_1=1, x2x_2, x3x_3, …\ldots be a sequence of real numbers such that for all n≥1n\geq 1 we have xn+1=xn+12xn. x_{n+1} = x_n + \frac 1{2x_n} . Prove that ⌊25x625⌋=625. \lfloor 25 x_{625} \rfloor = 625 .
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