$1 + 3 + 5 + \ldots + 99$

$\sum_0^{49}({1+2n}) = \sum_0^{49}(1)+ \sum_0^{49}(2n) = 50 + \frac{2\times49\times50}{2} = 2500$

I have used the fact that $\sum_1^n{k} = \frac{n(n+1)}{2}$

The sum of the first n odd numbers is aways equals to n^2 , 99 is the 50 odd number. So 50^2 = 2500

Let n be the nth odd number, then the sum is n^2. 99 is the 50th odd number so the answer is 2500.