We will find the solution to the following lhcc recurrence:an=6an-1-9an-2 for n\geqslant2 with initial conditions a0=2,a1=6. The first step as usual is to find the characteristic equation by trying a solution of the "geometric" format an=rn. (We assume also r\neq0). In this case we get:rn=6rn-1-9rn-2.Since we are assuming r?0 we can divide by the smallest power of r, i. e., rn-2 to get the characteristic equation:r2=6r-9.(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= ?Since the root is repeated, the general theory tells us that the general solution to our lhcc recurrence looks like:an=\alpha1(r)n+\alpha2n(r)nfor suitable constants \alpha1, \alpha2.To find the values of these constants we have to use the initial conditions a0=2,a1=6. These yield by using n=0 and n=1 in the formula above:2=\alpha1(r)0+\alpha20(r)0and 6=\alpha1(r)1+\alpha21(r)1By plugging in your previously found numerical value for r and doing some algebra, find \alpha1,\alpha2:\alpha1= ?\alpha2= ?Note the final solution of the recurrence is:an=\alpha1(r)n+\alpha2n(r)nwhere the numbers r,\alphai have been found by your work. This gives an explicit numerical formula in terms of n for the an.