Show that the sequence \{a_n\} is composed of terms of integers if a_0=1 and a_{n+1}=\dfrac{3a_n+\sqrt{5a^2_n-4} }{2}

Collected in the board: Sequence

Steven Zheng posted 2 days ago


(2a_{n+1}-3a_n)^2 =(\sqrt{5a^2_n-4})^2

4a^2_{n+1}-6a_na_{n+1}+9a^2_n = 5a^2_n-4

4a^2_{n+1} -6a_na_{n+1} +4a^2_n+4=0
4a^2_{n-1} -6a_na_{n-1} +4a^2_n+4=0

It's found that a_{n+1} and a_{n-1} are the two roots of the following quadratic equation

4x^2-6a_nx +4a^2_n+4 = 0

Using Vieta's formulas gives

a_{n+1}+a_{n-1} = 6a_n

Given that


a_2 = \dfrac{3a_1+\sqrt{5a^2_1-4} }{2} = 2

Using the equation (3), it can be deduced that all other terms of the sequence are also integers by using the method of mathematical induction.

Steven Zheng posted 2 days ago

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