If a sequence $$\{a_n\}$$ has its first term $$a_0=1$$,
$$a_{n+1}=\dfrac{7a_n+\sqrt{45{a_n}^2-36}}{2}$$, $$(n\in N)$$, show that

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If a sequence $$\{a_n\}$$ has its first term $$a_0=1$$, 
$$a_{n+1}=\dfrac{7a_n+\sqrt{45{a_n}^2-36}}{2}$$, $$(n\in N)$$, show that

Uniteasy Admin · Accepted Answer

Since the square root of any real number is always larger than or equal to zero,
$$a_{n+1}=\dfrac{7a_n+\sqrt{45{a_n}^2-36}}{2} \geq \dfrac{7a_n}{2}  >a_n$$
then the sequence $$\{a_n\}$$ is increasing. Given that $$a_0=1$$, $$a_n$$ is always positive.
Take the square of the equation after moving the terms of non square root to one side, and leave the term of square root at the other side.
$$\bigg(  a_{n+1}-\dfrac{7a_n}{2}\bigg)^2  = \bigg(  \dfrac{\sqrt{45a^2_n-36}  }{2}  \bigg)^2  $$
Expand and reorganize the terms, 
$$a^2_{n+1}-7a_{n+1}a_n+a^2_n+9=0$$n
Replacing $$n$$ with $$n-1$$, we get
$$a^2_{n-1}-7a_{n-1}a_n+a^2_n+9=0$$n
It's found that $$a_{n+1}$$ and $$a_{n-1}$$ are the two roots of quadratic equation
$$x^2-7a_nx +a^2_n+9 = 0$$
Using Vieta's formulas gives
$$a_{n+1}+a_{n-1} = 7a_n $$n
Since $$a_0 = 1$$,
$$a_1 = \dfrac{7a_0+\sqrt{45{a_0}^2-36}}{2} = 5$$
multiplication and addition of any integers yields integers according to the expression (3). Now we have proved that $$\{a_n\}$$ are positive integers From the expression (1) 
 
$$a^2_{n+1}+2a_{n+1}a_n-9a_{n+1}a_n+a^2_n+9=0  $$
$$ 9a_{n+1}a_n-9=(a_{n+1}+a_n)^2$$
$$a_{n+1}a_n-1 = (\dfrac{a_{n+1}+a_n}{3}  )^2$$,
Since both $$\{a_n\}$$ and $$\{a_{n+1}\}$$ are positive integers, 
$$a_na_{n-1}-1$$ are also positive integers and perfect square numbers.

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