Solve the recurrence relation a5n 3a5n 1 9a5n 2 27a5n 3

Solve the recurrence relation a^5_n + 3a^5_n -1 - 9a^5_n - 2 - 27a^5_n - 3 = 1 where n greaterthanorequalto 3 and a_0 = 0, a_1 = 1, and a_2 = 2.

given that an^5+3a(n-1)^5-9a(n-2)^5-27a(n-3)^5=1

Recurrence relation is an equation defines a sequance or multidimenssinoal array of values.

that is x_n+1=rx_n(1-xn)

the equation is a_n=c1 a(n-1)+c2a(n-2)+----+cda(n-d)

simple recurrence is an=ra(n-1)

here rⁿ=ar^n-1+br^n-2

given that an^5+3a(n-1)^5-9a(n-2)^5-27a(n-3)^5=1

==>an^5+3a^5((n-1)-3(n-2)-9(n-3))=1

==>an^5+3an^5=1

==>4an^5=1