The nth term of the sequence is given by \(2^{2n - 1}\). To find the first four terms, we substitute \(n = 1, 2, 3, \text{ and } 4\) into the expression:

\(2^{2(1) - 1} = 2^{1} = 2\)

\(2^{2(2) - 1} = 2^{3} = 8\)

\(2^{2(3) - 1} = 2^{5} = 32\)

\(2^{2(4) - 1} = 2^{7} = 128\)

Now, to find the sum of the first four terms, we add them together:

\(2 + 8 + 32 + 128 = 170\)