$9+11+13+15+\cdots+99=\sum_{k=0}^{45} \square$

Step 1 ：Define the first term (a) as 9, the last term (an) as 99, and the common difference (d) as 2.

Step 2 ：Calculate the number of terms (n) using the formula \(n = \frac{{an - a}}{d} + 1\). Substituting the values, we get \(n = \frac{{99 - 9}}{2} + 1 = 46.0\).

Step 3 ：Calculate the sum (S) of the series using the formula \(S = \frac{n}{2} \cdot (a + an)\). Substituting the values, we get \(S = \frac{46.0}{2} \cdot (9 + 99) = 2484.0\).

Step 4 ：Since the sum of an arithmetic series should be an integer, we can conclude that the sum is 2484.

Step 5 ：Final Answer: \(\boxed{2484}\)