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a_{n+1} =\sqrt{a_n^2 + 1}, a_{2n_0} = 3a_{n_0}, a_{46} wanted

Source: 2013 Saudi Arabia Pre-TST 4.1

September 13, 2020

Sequencealgebrarecurrence relation

Problem Statement

Let

a_1,a_2, a_3,...

be a sequence of real numbers which satisfy the relation

a_{n+1} =\sqrt{a_n^2 + 1}

Suppose that there exists a positive integer

n_0

such that

a_{2n_0} = 3a_{n_0}

. Find the value of

a_{46}