Answer to Question #117121 in Combinatorics | Number Theory for Priya

Question

Is the sequence 〖{a〗_n} a solution of the recurrence relation a_n=8a_(n-1)-16a_(n-2)if a_n=n^4 a_n=0.

Answer 1

Solution.

$a_n=8a_{n-1}-16a_{n-2};$

a) $a_n=0?$

Replace $n$ in $a_n=0$ by $n-1$

$a_{n-1}=0$;

Replace $n$ in $a_n=0$ by $n-2$

$a_{n-2}=0;$

$8a_{n-1}-16a_{n-2}=8\sdot0-16\sdot 0=0=a_n$; Yes

b)$a_n=n^4?$

Replase $n$ in $a_n=n^4$ by $n-1$

$a_{n-1}=(n-1)^4$;

Replase $n$ in $a_n=n^4$ by $n-2$

$a_{n-2}=(n-2)^4$

$8(n-1)^4-16(n-2)^4=8(n^4-4n^3+6n^2-4n+1)-16(n^4-8n^3+24n^2+-32n+16)=8n^4-32n^3+48n^2-32n+8-16n^4+128n^3-384n^2+512n-256=-8n^4+96n^3-336n^2+480n-248{=}\mathllap{/\,}n^4;$

No

Answer: a)Yes;

b)No.