Answer to Question #349270 in Discrete Mathematics for sara

Question #349270

Show that the explicit sequence {an} where an= 2n+1





-1 for n > or = to 1 is a





solution of the recurrence relation: an= 3an-1 – 2an-2 , n >or = to 3

1
Expert's answer
2022-06-09T14:00:57-0400

Given "a_n=2^{n+1}-1, n\\ge 1."

Then


"a_{n-1}=2^{n-1+1}-1=2^n-1"

"a_{n-2}=2^{n-2+1}-1=2^{n-1}-1"

Substitute


"3a_{n-1}-2a_{n-2}=3(2^n-1)-2(2^{n-1}-1)"

"=3(2^n)-3-2^n+2=2^{n+1}-1=a_n, n\\ge 3"

Since the last equation is true, then the explicit sequence "\\{a_n\\}" where "a_n=2^{n+1}-1, n\\ge 1" is a solution of the recurrence relation:


"a_n=3a_{n-1}-2a_{n-2}, n\\ge 3"


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