If your daughter is given a problem like this, then she probably knows, and can use, the formula for the sum of the first "n" numbers. That's one of the difficulties with you posting rather than her- we cannot ask directly.
If you or she does not know that formula, write out
1+ 2 + 3+ ...+ n-1+ n and reverse it
n+ n-1+ n-2+ ..._ 2 + 1 and add each pair vertically.
The sum of each pair of numbers is n+ 1 and there are n pairs so the sum of both sets of numbers is n(n+1). Finally, because there are two sums from 1 to n there, divide by 2: the sum of all integers from 1 to n is n(n+1)/2. The sum of all integers from 1 to 300 is (300(301)/2= 150(301)= 45150
we can find a formula for the sum of even numbers from 2 to 2n by writing 2+ 4+ 6 +...+ 2n-2+ 2n= 2(1)+ 2(2)+ 2(3)+ ...+ 2(n-1)+ 2n= 2(1+ 2+ 3+ ...+ n-1+ n)= 2(n(n+1)/2)= n(n+ 1). The sum of even integers from 1 to 300= 2(150) is 150(151)= 22650.
And since all numbers are either even or odd, we can find the sum of the first 150 odd integers, that is the sum of all odd integers from 1 to 299 by subtracting the sum of all even numbers from 2 to 300 from the set of all integers from 1 to 300: 45150- 22650= 22500.