The given series: (3, -6, +12, -24)

(a = 3, \\quad r = -2)

We can use the sum formula for a geometric series:

\(S = \\frac{a(1-r^n)}{1-r}\)

Substituting the values:

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

Simplifying further:

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

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

(-2^8 = -2^{n-1})

Since (2^{n-1}) is negative, we can drop the negative sign:

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

Since the bases are the same, the exponents must be equal:

\(8 = n-1\)

Solving for (n):

\(n = 9\)

Therefore, we need 9 terms of the series to make a total of (1-2^8).