(1 point) We will find the solution to the following lhcc recurrence: an = 8an_1 1611,14 for n z 2 with initial conditions a0 = 3,611 = 7. The first step as usual is to find the characteristic equation by trying a solution of the "geometric" format an = r". (We assume also r 7': 0). In this case we get: r" = 8r"_1 16r"_2. Since we are assuming r 7': 0 we can divide by the smallest power of r, i.e., r"'2 to get the characteristic equation: r2 = 8r 16. (Notice since our lhcc recurrence was degree 2, the characteristic equation is degree 2.) This characteristic equation has a single root r. (We say the root has multiplicity 2). Find r. r=D Since the root is repeated, the general theory (Theorem 4 in section 7.2 of Rosen) tells us that the general solution to our lhcc recurrence looks like: an = r110)" + a2n(r)\" for suitable constants a1, 052. To find the values of these constants we have to use the initial conditions do = 3, a1 = 7. These yield by using n=0 and n=1 in the formula above: 3 = am" + 0:200)" and 7 = a1(r)1+ 0521(r)1 By plugging in your previously found numerical value for r and doing some algebra, find (11, a2: a2 = Note the final solution of the recurrence is: an = r110)" + a2n(r)" where the numbers r, 05,- have been found by your work. This gives an explicit numerical formula in terms of n for the an