We will find the solution to the following Ihcc recurrence: an=-ban-1-9an-2 for n 2 with...

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We will find the solution to the following Ihcc recurrence: an=-ban-1-9an-2 for n ≥ 2 with initial conditions α = 3, a₁ = 6. 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 # 0). In this case we get: ¹-6-1-9-2 Since we are assuming r #0 we can divide by the smallest power of r, i.e., 2 to get the characteristic equation: 7-²=-6r-9. (Notice since our Ihcc 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 = Since the root is repeated, the general theory (Theorem 2 in section 5.2 of Rosen) tells us that the general solution to our Ihcc recurrence looks like: for suitable constants α1, 02. To find the values of these constants we have to use the initial conditions ao = 3, a₁ = 6. These yield by using n=0 and n=1 in the formula above: and an = α₁(r)" + α₂(r)n Note the final solution of the recurrence is: 3= a₁(r) + a20(r)⁰ 6 = α₁(r)¹ + a₂¹(r)¹ By plugging in your previously found numerical value for r and doing some algebra, find α₁, 02: α₁ = α2 = α₁(r)¹ + α(r)n where the numbers r, a; have been found by your work. This gives an explicit numerical formula in terms of n for the an an = We will find the solution to the following Ihcc recurrence: an=-ban-1-9an-2 for n ≥ 2 with initial conditions α = 3, a₁ = 6. 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 # 0). In this case we get: ¹-6-1-9-2 Since we are assuming r #0 we can divide by the smallest power of r, i.e., 2 to get the characteristic equation: 7-²=-6r-9. (Notice since our Ihcc 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 = Since the root is repeated, the general theory (Theorem 2 in section 5.2 of Rosen) tells us that the general solution to our Ihcc recurrence looks like: for suitable constants α1, 02. To find the values of these constants we have to use the initial conditions ao = 3, a₁ = 6. These yield by using n=0 and n=1 in the formula above: and an = α₁(r)" + α₂(r)n Note the final solution of the recurrence is: 3= a₁(r) + a20(r)⁰ 6 = α₁(r)¹ + a₂¹(r)¹ By plugging in your previously found numerical value for r and doing some algebra, find α₁, 02: α₁ = α2 = α₁(r)¹ + α(r)n where the numbers r, a; have been found by your work. This gives an explicit numerical formula in terms of n for the an an =