I'm not sure what the question means by "use S = n^2"...
At any rate, the given sum
1 + 3 + 5 + … + 301
is simply the sum of the first 151 odd positive integers, which we can write in the form 2n - 1 for n between 1 and 151. That is,
[tex]1+3+5+\cdots+301 = \displaystyle\sum_{n=1}^{151}(2n-1)[/tex]
We can expand the sum as
[tex]\displaystyle 2\sum_{n=1}^{151}n - \sum_{n=1}^{151}1[/tex]
Next, recalling that
[tex]\displaystyle \sum_{i=1}^n1 = 1+1+1+\cdots+1 = n \text{ and }\sum_{i=1}^ni=1+2+3+\cdots+n = \frac{n(n+1)}2[/tex]
it follows that
[tex]1+3+5+\cdots+301 = 2\cdot\dfrac{151\cdot152}2 - 151 = 151\cdot(152-1) = 151^2 = \boxed{22,801}[/tex]
(Probably "S = n^2" means "sum of n odd integers = n ²")
