Question

Solve the recurrence relation a_n=-3a_(n-1)-3a_(n-2)-a_(n-3) with a_0=5,a_1=-9,and a_2=15.

Answer 1

Solution.

$a_n=-3a_{n-1}-3a_{n-2}-a_{n-3}, a_0=5, a_1=-9, a_2=15;$

Let $a_n=r^3, a_{n-1}=r^2, a_{n-2}=r, a_{n-3}=1,$then

$r^3=-3r^2-3r-1;$

$r^3+3r^2+3r+1=0$;

$(r+1)^3=0;$

$r+1=0;$

$r=-1;$

$a_n=\alpha_1r^n+\alpha_2nr^n+ … +\alpha_kn^k-1r^n;$

So r=-1, then $a_n=\alpha_1(-1)^n+\alpha_2n(-1)^n+\alpha_3n^2(-1)^n;$

Substitute the value and find $\alpha_1,\alpha_2,\alpha_3:$

$5=\alpha_1(-1)^0, \alpha_1=5;$

$-9=\alpha_1(-1)^1+\alpha_2(-1)^1+\alpha_3(-1)^1=-\alpha_1-\alpha_2-\alpha_3;$

$15=\alpha_1(-1)^2+\alpha_2\sdot2\sdot(-1)^2+\alpha_3\sdot4\sdot(-1)^2=\alpha_1+2\alpha_2+4\alpha_3;$

$-9=-5-\alpha_2-\alpha_3;$ | $\sdot2$

$15=5+2\alpha_2+4\alpha_3;$

$-18=-10-2\alpha_2-2\alpha_3;$

$15=5+2\alpha_2+4\alpha_3;$

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$-3=-5+2\alpha_3;$

$2=2\alpha_3;$

$\alpha_3=1;$

$15=5+2\alpha_2+4;$

$6=2\alpha_2;$

$\alpha_2=3;$

$a_n=5(-1)^n+3n(-1)^n+1n^2(-1)^n=(-1)^n(5+3n+n^2);$

Answer: $a_n=(-1)^n(5+3n+n^2).$