Let $$[x]$$ denote greatest integer less than or equal to $$x$$. If for $$n \in N$$, $$\left(1 - x + x^3\right)^n = \sum_{j=0}^{3n} a_j x^j$$, then $$\sum_{j=0}^{\left[\frac{3n}{2}\right]} a_{2j} + 4\sum_{j=0}^{\left[\frac{3n-1}{2}\right]} a_{2j+1}$$ is equal to:

We are given $$(1 - x + x^3)^n = \sum_{j=0}^{3n} a_j x^j$$ and need to find $$\sum_{j=0}^{[3n/2]} a_{2j} + 4\sum_{j=0}^{[(3n-1)/2]} a_{2j+1}$$.

Substituting $$x = 1$$: $$(1 - 1 + 1)^n = 1 = \sum_{j=0}^{3n} a_j$$. This gives $$\sum a_{2j} + \sum a_{2j+1} = 1$$.

Substituting $$x = -1$$: $$(1 + 1 - 1)^n = 1 = \sum_{j=0}^{3n} a_j(-1)^j$$. This gives $$\sum a_{2j} - \sum a_{2j+1} = 1$$.

Adding these two equations: $$2\sum a_{2j} = 2$$, so $$\sum a_{2j} = 1$$. Subtracting: $$2\sum a_{2j+1} = 0$$, so $$\sum a_{2j+1} = 0$$.

Therefore, the required expression equals $$\sum a_{2j} + 4\sum a_{2j+1} = 1 + 4(0) = 1$$.