For each part of this problem a difference equation and one or more initial values are given. use difference equation to compute additional terms as specified

(a) if a0 = 10 and an+1 = 4+0.6an, compute a1, a2 and a3.

Well, let's see . . .

When n=0, A(n) = 10

Also A(0+1) = A(1) = 4 + 0.6(A0)
                                  = 4 + 0.6(10)
                                  = 4  +  6        =    10

When n=1,   A(1+1) = A(2) = 4 + 0.6(A1)
                                              = 4 + 0.6(10)
                                              = 4  +  6         =     10

When n=2,  A(2+1) = A(3) = 4 + 0.6(A2)
                                             = 4 + 0.6(10)
                                             = 4  +  6           =     10 .

Amazing !  Whoda thunk it ? !

By my calculations ... which are often bullet-proof and air-tight,
but are just as often worthless and absurd ...

A1,  A2,  A3,  and all other higher terms, are all  10 . 
 

Hey bby was a good way i just got