Question

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

\{a_n\} are positive integers

a_na_{n-1}-1 are perfect square numbers.

Steven Zheng posted 1 day ago

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
(1)

Replacing n with n-1, we get

a^2_{n-1}-7a_{n-1}a_n+a^2_n+9=0
(2)

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
(3)

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.

Steven Zheng posted 1 day ago

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