Let $[x]$ denote greatest integer less than or equal to $x$. If for $n \in N,\left(1-x+x^{3}\right)^

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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

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(b) : We have, $\left(1-x+x^{3}ight)^{n}$ $$\begin{equation*}=a_{0} x^{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots+a_{3 n} x^{3 n} 	ag{i} \end{equation*}$$ Putting $x=1$ in (i), we get $$\begin{equation*}a_{0}+a_{1}+a_{2}+\ldots \ldots+a_{3 n}=1 	ag{ii} \end{equation*}$$ Also, putting $x=-1$ in (i), we get $$\begin{equation*}a_{0}-a_{1}+a_{2}-a_{3}+\ldots \ldots+(-1)^{3 n} \cdot a_{3 n}=1 	ag{iii} \end{equation*}$$ Now, adding and subtracting (ii) and (iii), we get $2\left\{a_{0}+a_{2}+a_{4}+\ldots . .+a_{2}\left[\frac{3 n}{2}ight]ight\}=2$ $$\begin{equation*}\Rightarrow \sum_{j=0}^{[3 n / 2]} a_{2 j}=1 	ag{iv} \end{equation*}$$ and $2\left\{a_{1}+a_{3}+a_{5}+\ldots \ldots+a_{2}\left[\frac{3 n-1}{2}ight]+1ight\}=0$ $$\begin{equation*}\Rightarrow \quad\left[\frac{3 n-1}{2}ight] a_{2 j+1}=0 	ag{v} \end{equation*}$$ $	herefore \quad \sum_{j=0}^{[3 n / 2]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}ight]} a_{2 j+1}=1+4 	imes 0$\ (Using (iv) and (v))\ $=1$\