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a_{n+1}=2a_n/(1+a_n^2), c_{n+1}=c_n^2-2c_n+2, c_n=2c_0c_1...c_{n-1}/c_n

Source: Austrian - Polish 1994 APMC

May 3, 2020
recurrence relationSequencealgebra

Problem Statement

The sequences (an)(a_n) and (c_n) are given by a0=12a_0 =\frac12, c0=4c_0=4 , and for n≥0n \ge 0 , an+1=2an1+an2a_{n+1}=\frac{2a_n}{1+a_n^2}, cn+1=cn2−2cn+2c_{n+1}=c_n^2-2c_n+2 Prove that for all n≥1n\ge 1, an=2c0c1...cn−1cna_n=\frac{2c_0c_1...c_{n-1}}{c_n}
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