Let $[x]$ denote greatest integer less than or equal to $x$. If for $n \in \mathbb{N},\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$, then $\left.\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right.}\right] a_{2 j}+1$ is equal to :
Correct Option: , 3
$\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$
$\left(1-x+x^{3}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2} \ldots \ldots+a_{3 n} x^{3 n}$
$\sum_{j=0}^{\frac{3 n}{2}} a_{2 j}=\operatorname{Sum}$ of $a_{0}+a_{2}+a_{4} \ldots \ldots$
$\left[\frac{3 n-1}{2}\right]$
$\sum_{j=0} a_{2 j}+1=$ Sum of $a_{1}+a_{3}+a_{5} \ldots \ldots$
put $x=1$
$1=a_{0}+a_{1}+a_{2}+a_{3} \ldots \ldots \ldots+a_{3 n} \ldots \ldots(A)$
Put $x=-1$
$1=a_{0}-a_{1}+a_{2}-a_{3} \ldots \ldots \ldots+(-1)^{3 n} a_{3 n} \ldots \ldots$
Solving (A) and (B)
$a_{0}+a_{2}+a_{4} \ldots .=1$
$a_{1}+a_{3}+a_{5} \ldots \ldots=0$
$\left.\left.\sum_{j=0}^{\left[\frac{3 n}{2}\right.}\right] a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right.}\right] a_{2 j+1}=1$