How to describe an algorithm whose input size diminishes by 1 for each iteration
On recursive algorithms, you need first to obtain the recurrence relation. Then you solve the recurrence to get the total time required by the algorithm.

Let $n$ be the input size. Let $f(n)$ be the time required in each recursive call. Let $b$ be the input size for the base case. A general recurrence relation might look like this  

$$
T(n) = T(n-1) + f(n)  \quad \quad \textrm{if } n &gt; b \\
T(n) = c \quad \quad  \quad \quad \quad \quad \quad \quad \;\ \textrm{if } n = b 
$$
where $c$ typically is a constant (because typically the base case takes constant time).

So, you need to figure it out the function $f(n)$, which is specific for each algorithm. For example, if $f(n)=n$ (e.g. a simple pass on a array), and assuming $b=1$ and $c=1$ (it is not a problem to assume $c=1$ if you are working with assymptotic notation) then

$$
T(n) = T(n-1) + n  \quad \quad \textrm{if } n &gt; 1 \\
T(n) = 1 \quad \quad  \quad \quad \quad \quad \quad  \textrm{if } n = 1
$$

Solving this recurrence, you have that $$T(n) = n + (n-1) + \ldots + 2 + 1 = \frac{n(n+1)}{2} = O(n^2)$$.