Solving Recurring Relations Can you please help, my son has been trying for over two hours now to solve the following:
A sequence of terms $\left\{u_n\right\}$ is defined for $n\geq 1$, by the recurrence relation:
$$u_{n+1}=u_n/3.$$
Find $u_2$, $u_3$, $u_4$ and $u_5$ in terms of $u_1$.
Thank you
 A: Hint:  You just need to plug into the formula.  So $u_2=\frac {u_1}3$.  Now  $u_3=\frac {u_2}3=?$
A: $u_2 = \frac {u_1}{3}$  
$u_3 = \frac {u_1}{9}$  
$u_4 = \frac {u_1}{27}$  
$u_5 = \frac {u_1}{27 \times 3 = 81}$  
A: Since your son (and thus you...) might face similar exercises in the future it will not hurt to state the general idea.
You have some sequence $u_n$ defined not by a formula, but by some kind of relationship between nearby values.   In this question, the relationship is between adjacent values.  The specification "$u_{n+1}=u_{n}/3$" means you have a collection of equations, one for each $n=1,2,3,4...$, that starts $u_2 = u_1 / 3$, then $u_3 = u_2/3$ and so on. 
All these equations let you compute the next term from previous terms, compiling a list of the values of $u_1, u_2, u_3, \ldots$ one by one in order.  The important thing is that at every step the "next" term you would like to compute, can be "solved for", using one of the equations, from some of the earlier already-computed values in the sequence.  For example we can calculate $u_3$ from $u_2$ in this case, and every $u_n$ can be determined from the one before, except a few initial values (in this case, $u_1$) for which there is no equation relating them to previous terms of the sequence.   
The initial values determine all the later terms of the sequence, and are usually an external specification that determines, from the many different sequences that satisfy the recurrence relation, which particular one is being considered.  We don't solve for them (in most cases), we state what they are and that uniquely defines the rest of the sequence.
In some simple or lucky cases the recurrence relation can be converted into a formula that takes $n$ and directly writes down $u_n$ without computation of all earlier terms.  This is not something to expect as a possibility for more complicated recursions.
