Let [x] denote greatest integer less than or

$\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$

$\left(1-x+x^{3}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2} \ldots \ldots+a_{3 n} x^{3 n}$

$\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}=$ Sum of $a_{0}+a_{2}+a_{4} \ldots \ldots . .$

$\sum_{j=0}^{\left[\frac{3 \mathrm{n}-1}{2}\right]} \mathrm{a}_{2 \mathrm{j}}+1=$ Sum of $\mathrm{a}_{1}+\mathrm{a}_{3}+\mathrm{a}_{5} \ldots \ldots . .$

put $x=1$

$1=a_{0}+a_{1}+a_{2}+a_{3} \ldots \ldots \ldots+a_{3 n} \quad \ldots \ldots .(A)$

Put $x=-1$

$1=a_{0}-a_{1}+a_{2}-a_{3} \ldots \ldots \ldots+(-1)^{3 n} a_{3 n} \ldots \ldots . .(B)$

Solving (A) and (B)

$a_{0}+a_{2}+a_{4} \ldots . .=1$

$a_{1}+a_{3}+a_{5} \ldots \ldots . .=0$

$\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right]} a_{2 j+1}=1$