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

Question #117121
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.
1
Expert's answer
2020-06-01T17:11:47-0400

Solution.

an=8an116an2;a_n=8a_{n-1}-16a_{n-2};

a) an=0?a_n=0?

Replace nn in an=0a_n=0 by n1n-1

an1=0a_{n-1}=0;

Replace nn in an=0a_n=0 by n2n-2

an2=0;a_{n-2}=0;

8an116an2=80160=0=an8a_{n-1}-16a_{n-2}=8\sdot0-16\sdot 0=0=a_n; Yes

b)an=n4?a_n=n^4?

Replase nn in an=n4a_n=n^4 by n1n-1

an1=(n1)4a_{n-1}=(n-1)^4;

Replase nn in an=n4a_n=n^4 by n2n-2

an2=(n2)4a_{n-2}=(n-2)^4

8(n1)416(n2)4=8(n44n3+6n24n+1)16(n48n3+24n2+32n+16)=8n432n3+48n232n+816n4+128n3384n2+512n256=8n4+96n3336n2+480n248=/n4;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.

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