Update 2-3.md

replace i-the with i-th in c's answer.

docs/Chap02/Problems/2-3.md

@@ -45,7 +45,7 @@ The running time is $\Theta(n^2)$, because of the nested loop. It is obviously s
 
 **c.** **Initialization:** It is pretty trivial, since the summation has no terms which implies $y = 0$.
 
-**Maintenance:** By using the loop invariant, in the end of the $i$-the iteration, we have
+**Maintenance:** By using the loop invariant, in the end of the $i$-th iteration, we have
 
 $$
 \begin{aligned}