Let $n \in \mathbb Z^+ = \{1,2,3,\dots\}$. How many $(a,b) \in \mathbb Z^+ \times \mathbb Z^+$ are such that, $(a,b)S(n,n)$ where
$$(a,b)S(c,d) \iff a + b ≤ c + d$$
In our case, $(a,b)S(n,n) \iff a + b ≤ 2n$
I have found that for $n = 1$, then there is only one ordered pair, which is $(1,1)$.
For $n = 2$, there are $6$ ordered pairs.
For $n = 3$, there are $15$ ordered pairs.
I can't seems to find the pattern. Is my approche to this questions correct?
First let me correct your terminology: for $n=2$, for instance, what you mean is that there are $6$ ordered pairs that are in the relation $S$ to $(2,2)$. The objects $(1,1),(1,2),(2,1),(2,2),(1,3)$, and $(3,1)$ that you found are ordered pairs, not relations.
HINT: For a given $n$ you’re counting the pairs $(a,b)$ such that $a+b\le 2n$. One way to do that is to count the pairs with each possible sum up through $2n$ and add up the results. The smallest possible sum is $2$, from the pair $(1,1)$, and that’s the only pair with sum $2$. Let’s make a table showing the pairs with a given sum and see if any pattern emerges:
$$\begin{array}{c|l} n&\text{pairs with that sum}\\ \hline 2&(1,1)\\ 3&(1,2),(2,1)\\ 4&(1,3),(2,2),(3,1)\\ 5&(1,4),(2,3),(3,2),(4,1) \end{array}$$