Question

The sequence ${\left(x_n\right)}_{n\ge 1}.$ is definced by $x_1=1,x_2=3$ and $x_{n+2}=6x_{n+1}-x_n,$ for all $n\ge 1.$ Then $x_2+{(-1)}^n$ is a perfect square, for all :

• A. $n\ge 1.$
• B. $n\ge 2$
• C. $n\le 1$
• D. None of These

Answer 1

The correct option is A $n\ge 1.$

Let ,$y_n=x_n+{(-1)}^n.$
By inspection we find that $y_1=0,y_2=4,y_3=16,y_4=100,y_5=576$
etc, denote by,$Z_n=\sqrt{y_n};$
then,$Z_1=0,Z_2=2,Z_3=4,Z_4=10,Z_5=24$
and so on.The first terms of the sequence ${\left(Z_n\right)}_{n\ge 1}$
suggesting that$Z_{n+2}=2Z_{n+1}{Z}_n,$
for all $n\ge 1$
Indeed,we will prove by induction the following statement:$y_n=Z_n^2$
where $z_1=0,z_2=2$ and $z_{n+2}=2z_{n+1}{z}_n,$
for all $n\ge 1.$
The base case is easy to deal with. In order to prove the inductive step, we need a recursive equation for the sequence ${\left(y_n\right)}_{n\ge 1}.$
Since $x_n=y_n-{(-1)}^n,$
we have $y_{n+2}-{(-1)}^{n+2}=6y_{n+1}-6{(-1)}^{n+1}-y_n+{(-1)}^n,$
and also $y_{n+2}-{(-1)}^{n+1}=6y_n-6{(-1)}^n-y_{n-1}+{(-1)}^{n-1}.$
Adding up yields $y_{n+2}+y_{n+1}=6y_{n+1}+6y_n-y_n-y_{n-1},$ or,
$y_{n+2}=5y_{n+1}+5y_n-y_{n-1},$
for all $n\ge 2.$
Now assume, $y_k=z_k^2$,for $k=1,2,...,n+1.$
$y_{n+2}=5z_{n+1}^2+5z_n^2-z_{n-1}^2=5z_{n+1}^2+5z_n^2-{(z_{n+1}^2-2z_n^2)}^2=4z_{n+1}^2+z_n^2+4z_n+1z_n={(2z_{n+1}+z_n)}^2=z_{n+2}^2,$
and we are done. Observation The informed reader may notice that an alternative is possible if we write $x_n$ in closed form. Indeed, it is known that if and ,with$$, are the roots of the quadratic equation $x^2=ax+b$ then any sequence satisfying
$x_{n+2}=a{x}_{n+1}+b{x}_n$,for all $n\ge 1,$ has the form $x_n=c_1{\alpha }^n+c_2{\beta }^n,$ where the constants $c_1$ and $c_2$ can be determined from the first two terms of the sequence. In our case, the roots of the equation $x^2=6x-1$ are
$\alpha =3+2\sqrt{2},\beta =3-2\sqrt{2}$ ,hence, $x_n=c_1{(3+2\sqrt{2})}^n+c_2{(3-2\sqrt{2})}^n.$
Since $x_1=1$ and $x_2=3$ we obtain,
$c_1=\frac{1}{2}(3-2\sqrt{2}),c_1=\frac{1}{2}(3+2\sqrt{2}),x_n=\frac{1}{2}({(3\mid +2\sqrt{2})}^{n-1}+{(3-2\sqrt{2})}^{n-1}),$
for all $n\ge 1.$
It follows that $x_n+{(-1)}^n=\frac{1}{2}({(3+2\sqrt{2})}^{n-1}+{(3-2\sqrt{2})}^{n-1}+2{(-1)}^n)=\frac{1}{2}({(1+\sqrt{2})}^{2(n-1)}+{(1-\sqrt{2})}^{2(n-1)}-2{(-1)}^{n-1})$
$={\left(\frac{{(1+\sqrt{2})}^{n-1}-{(1-\sqrt{2})}^{n-1}}{\sqrt{2}}\right)}^2.$
The sequence $w_n=\frac{{(1+\sqrt{2})}^{n-1}-{(1-\sqrt{2})}^{n-1}}{\sqrt{2}}$ has the form $w_n=c_1{\alpha }^n+c_2{\beta }^n,$ with $\alpha =1+\sqrt{2},\beta =1-\sqrt{2},$
hence it satisfies the recursive equation $w_{n+2}=2w_{n+1}+w_n,$
for all $n\ge 1.$
Finally,since $w_1=0$ and $w_1=2$ we deduce that
$w_n$ is an integer for all $n\ge 1.$ $