Let x denote greatest integer less than or equal to x . If for n ∈ N , 1 - x + x 3 n = ∑ j = 0 3…

${\left(1-x+x^3\right)}^n=\sum _{j=0}^{3n}{a}_j{x}^j$

$\Rightarrow {\left(1-x+x^3\right)}^n=a_0+a_{1}x+a_2{x}^2\dots \dots +a_{3n}{x}^{3n} ...\left(i\right)$

Then,

$\sum _{j=0}^{\left[\frac{3n}{2}\right]}{a}_{2j}=a_0+a_2+a_4\dots \dots$ 

$\sum _{j=0}^{\left[\frac{3n-1}{2}\right]}{a}_{2j+1}=a_1+a_3+a_5\dots \dots$

Put $x=1$ in $\left(i\right)$, then we get

$1=a_0+a_1+a_2+a_3\dots \dots \dots +a_{3n}$   $...\left(\mathrm{A}\right)$

Put $x=-1$ in $\left(i\right)$, then we get

$1=a_0-a_1+a_2-a_3\dots \dots \dots +{\left(-1\right)}^{3n}{a}_{3n}$    $...\left(\mathrm{B}\right)$

Solving $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right)$

$a_0+a_2+a_4\dots ..=1$

$a_1+a_3+a_5\dots \dots =0$

Hence,

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