Since there are only 4 terms in the sum, it's not too much work to expand it as
[tex]\displaystyle \sum_{n=0}^3 (2n-3) = -3 + (-1) + 1 + 3 = \boxed{0}[/tex]
Alternatively, we can use the well-known formulas
[tex]\displaystyle \sum_{n=1}^N 1 = \underbrace{1 + 1 + \cdots + 1}_{N \text{ 1s}} = N[/tex]
[tex]\displaystyle \sum_{n=1}^N n = 1+2+\cdots+N = \frac{N(N+1)}2[/tex]
These sums start at n = 0, so in our given sum we will keep track of the 0-th term separately:
[tex]\displaystyle \sum_{n=0}^3 (2n-3) = -3 + \sum_{n=1}^3 (2n-3) = -3 + 2 \sum_{n=1}^3 n - 3 \sum_{n=1}^3 1 = -3 + 3\times4 - 3\times3 = 0[/tex]
as expected.
