Solve the simultaneous recurrence relations

Question

an = an−1 + bn−1
bn = an−1 − bn−1
with a0 = 1 and b0 = 2.

Answer 1

$a_{n}=a_{n-1}+b_{n-1}$     ……………………………...(1)

$b_{n}=a_{n-1}-b_{n-1}$  ……………………………………..(2)

doing $(1)+(2) $ ,

$b_{n}+a_{n}=2a_{n-1}$  

So , $b_{n-1}+a_{n-1}=2a_{n-2}$

 or ,$b_{n-1}=2a_{n-2}-a_{n-1}$

Substituting value of $b_{n-1}$ in (1) we get ,

$a_{n}=a_{n-1}+2a_{n-2}-a_{n-1}$

or, $a_{n}=2a_{n-2}$

Solving ,linear homogeneous equation we get ,

$r^{n}=2r^{n-2}$

or, $r^{2}-2=0$

So ,$r=\pm \sqrt{2}$

So,$a_{n}=\alpha 1\left ( \sqrt{2} \right )^{n}+\alpha 2\left ( -\sqrt{2} \right )^{n}$,

Now given , $a_{0}=1$,$a_{1}=3$,

So ,$\alpha 1$+$\alpha 2=1$ …..…..(3)

     $\sqrt{2} \alpha 1-\sqrt{2}\alpha 2=3$…...……..(4)

So , $\alpha 1=\frac{\sqrt{2}+3}{2\sqrt{2}}$,$\alpha 2=\frac{\sqrt{2}-3}{2\sqrt{2}}$,

So, $a_{n}=\frac{\sqrt{2}+3}{2\sqrt{2}}(\sqrt{2})^{n}+\frac{\sqrt{2}-3}{2\sqrt{2}}(-\sqrt{2})^{n}$

$b_{n}=2a_{n}-a_{n}$ you can easily found by substituting $a_{n}$ .