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Let $[ x ]$ denote greatest integer less than or equal to $x .$ If for $n \in N ,\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$, then $\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right]} a_{2 j+1}$ is equal to
$2$
$2^{ n -1}$
$1$
$n$
Solution
$\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$
$\left(1-x+x^{3}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2} \ldots \ldots .+a_{3 n} x^{3 n}$
$\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}=\operatorname{Sum}$ of $a_{0}+a_{2}+a_{4} \ldots \ldots . .$
$\sum_{j=0}^{\left[\frac{3 n -1}{2}\right]} a_{2 j+1} =$ Sum of $a_{1}+a_{3}+a_{5} \ldots \ldots$
put $x=1$
$1= a _{0}+ a _{1}+ a _{2}+ a _{3} \ldots \ldots \ldots+ a _{3 n } \quad \ldots \ldots$
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)
$a_{0}+a_{2}+a_{4} \ldots . .=1$
$a_{1}+a_{3}+a_{5} \ldots \ldots .=0$
$\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right]} a_{2 j+1}=1$
