We will find the solution to the following lhcc recurrence: an=6an19an2 for n2 with initial conditions a0=3,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=0 ). In this case we get: rn=6rn19rn2 Since we are assuming r=0 we can divide by the smallest power of r, i.e., rn2 to get the characteristic equation: r2=6r9 (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 (Theorem 2 in section 5.2 of Rosen) tells us that the general solution to our Ihcc recurrence looks like: an=1(r)n+2n(r)n for suitable constants 1,2 To find the values of these constants we have to use the initial conditions a0=3,a1=6. These yield by using n=0 and n=1 in the formula above: 3=1(r)0+20(r)0 and 6=1(r)1+21(r)1 By plugging in your previously found numerical value for r and doing some algebra, find 1,2. 1= 2= Note the final solution of the recurrence is: an=1(r)n+2n(r)n where the numbers r,i have been found by your work. This gives an explicit numerical formula in terms of n for the an