We will find the solution to the following Ihcc recurrence a

We will find the solution to the following Ihcc recurrence: an= -2an-1 + 3an-2 for n > 2 with initial conditions a0 = 3, a1 =8. The first step in any problem like this is to find the characteristic equation by trying a solution of the \"geometric\" formatan=rn. (We assume also r 0). In this case we get: rn= -2rn-1+3rn-2 Since we are assuming r Owe can divide by the smallest power of r, i.e., rn-2to get the characteristic equation: r2 = -2r + 3. [Notice since our Ihcc recurrence was degree 2, the characteristic equation is degree 2] Find the two roots of the characteristic equation r\\ and r2. When entering your answers use r1 le r2: r1= r2= Since the roots are distinct, the general theory {Theorem 1 in Section 8.2 of Rosen) tells us that the general solution to our Ihcc recurrence looks like: an- alpha (r1)n+alpha 2(r2)n for suitable constants alpha1, alpha2 To find the values of these constants we have to use the initial conditions a0=3,a1=8. These yield by using n = 0 and n = 1 in the formula 3 = alpha1(r1)n+alpha2(r2)n By plugging in your previously found numerical values for r\\ and r2 and doing some algebra, find alph1,alpha2 {Be careful to note that (-x)n (xn) when n is even, for example (-3)2 -(32).) Alpha1= alpha2=

Solution

A[N]=-2A[N-1]+3A[N-2] A0=3