We will find the solution to the following lhcc recurrence: an : 1Gm1 + 2an_2 for n 2 2 with initial conditions a0 : 1, a1 : 6. The first step in any problem like this is to find the characteristic equation by trying a solution of the "geometric" format 0.71 = r". (We assume also 1' 79 0). In this case we get: 1*" : 1r"_1 + 2r"_2. Since we are assuming 7' 73 0 we can divide by the smallest power of r, i.e., r"'2 to get the characteristic equation: 1'2 : 1'r + 2. (Notice since our lhcc recurrence was degree 2, the characteristic equation is degree 2.) Find the two roots of the characteristic equation 11 and 7'2. When entering your answers use r1 3 7'2: 7'1: .72: Since the roots are distinct, the general theory (Theorem 1 in section 5.2 of Rosen) tells us that the general solution to our lhcc recurrence looks like: an : 0:101)" + 0:202)" for suitable constants a1, (12. To find the values of these constants we have to use the initial conditions an : 1, a1 : 6. These yield by using n:0 and n=1 in the formula above: 1 : (11(1'1)0 + a2(r2)0 and 6 : 01(7'1)1 + Ctg('l"2)1 By plugging in your previously found numerical values for 1'1 and 7'2 and doing some algebra, find a1, a2: [Be careful to note that (33)" y (m") when n is even, for example (3)2 y (32).] Cu : $ on; : Note the final solution of the recurrence is: an : a1(r1)" + 02W)" where the numbers n, ai have been found by your work. This gives an explicit numerical formula in terms of n for the an