[tex]\displaystyle \sum_{n=1}^{15} (2n-1) = 2 \sum_{n=1}^{15} n - \sum_{n=1}^{15}1[/tex]
Recall that
[tex]\displaystyle\sum_{n=1}^N 1 = N[/tex]
[tex]\displaystyle \sum_{n=1}^N n = \dfrac{N(N+1)}2[/tex]
Then
[tex]\displaystyle \sum_{n=1}^{15} (2n-1) = 2\cdot\dfrac{15\cdot16}2 - 15 = \boxed{225}[/tex]
Alternatively, notice that the terms in the sum are odd integers, and terms on opposite ends of the sum add up to 30, except the middle term:
1 + 3 + 5 + 7 + … + 23 + 25 + 27 + 29
= (1 + 29) + (3 + 27) + (5 + 25) + … + (13 + 17) + 15
= [30 + 30 + 30 + … + 30] + 15
= 7×30 + 15
= 210 + 15
= 225
