In other words, they "suggest" that the general formula is U_{n}= 3n- 2[/sub].

All you need to do is show that this does, in fact, satisfy the recursion.

If U_{n}= 3n- 2, then U_{k-1}= 3(k- 1)- 2= 3k- 5, U_{k}= 3k- 2 and it is certainly true that U_{k}= 3k- 2= 3k- 5+ 3= U[sub]k-1]+ 3