Section 53 81 82 Problem 3 1 pt Find the solution to the fol

Section 5.3 8.1 8.2: Problem 3 (1 pt) Find the solution to the following Ihcc recurrence: an= 81an-2 for n geq 2 with the initial condition a0 =-4, a1 = 18.

Solution

a0 = -4

a1 =18

Hence a2 = 81a0 = -324

a3 = 81 a1 = 1458

Thus an if n is even is a term in the Geometric series with common ratio 81 and first term -4

where as if n is odd, it is a term in the Geometric series with common ratio 81 and the first term 18

Thus a1, a3, a5,,,,,,= 18, 18(81), 18(81)², .....

and a0, a2, a4 ,,,, = -4, -4(81), -4(81)².....

Let n be of the form 2m+1 if odd and 2m if even.

an = -4(81)^m if n = 2m

= 18(81)^m if n = 2m+1