Let [x] denote greatest integer less than or equal to x . If for n...

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

$⇒ {\left(1 - x + {x}^{3}\right)}^{n} = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} \ldots \ldots + {a}_{3n}{x}^{3n} . . . \left(i\right)$

Then,

$\mathop{∑}_{j = 0}^{\left[\dfrac{3n}{2}\right]}{a}_{2j} = {a}_{0} + {a}_{2} + {a}_{4} \ldots \ldots$

$\mathop{∑}_{j = 0}^{\left[\dfrac{3n - 1}{2}\right]}{a}_{2j + 1} = {a}_{1} + {a}_{3} + {a}_{5} \ldots \ldots$

Put x = 1 in \left(i\right)$, then we get</p><p>$1 = {a}_{0} + {a}_{1} + {a}_{2} + {a}_{3} \ldots \ldots \ldots + {a}_{3n} . . . ( A )$</p><p>Put x = - 1 in \left(i\right)$, then we get

$1 = {a}_{0} - {a}_{1} + {a}_{2} - {a}_{3} \ldots \ldots \ldots + {\left(- 1\right)}^{3n}{a}_{3n} . . . ( B )$

Solving $( A ) and ( B )$

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

${a}_{1} + {a}_{3} + {a}_{5} \ldots \ldots = 0$

Hence,

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