Let [ x ] denote greatest integer less than or equal to x. If for n$\in$N,

${(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}}$,

then $$\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}} + 4} \sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j}} + 1} $$ is equal to :

${(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}}$<br><br>$(1 - x + {x^3}) = {a_0} + {a_1}x + {a_2}{x^2} + ...... + {a_{3n}}{x^{3n}}$<br><br>Put x = 1<br><br>$1 = {a_0} + {a_1} + {a_2} + {a_3} + {a_4} + ........ + {a_{3n}}$ ...... (1)<br><br>Put x = $-$1<br><br>$1 = {a_0} - {a_1} + {a_2} - {a_3} + {a_4} + ........( - 1){}^{3n}{a_{3n}}$ ..... (2)<br><br>Add (1) + (2)<br><br>$\Rightarrow {a_0} + {a_2} + {a_4} + {a_6} + ...... = 1$<br><br>Sub (1) $-$ (2)<br><br>$\Rightarrow {a_1} + {a_3} + {a_5} + {a_7} + ...... = 0$<br><br>Now, $$\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}}} + 4\sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j }}} + 1 $$<br><br>$= ({a_0} + {a_2} + {a_4} + ......) + 4({a_1} + {a_3} + .....)$<br><br>$= 1 + 4 \times 0$ + 1<br><br>$= 1 + 1 = 2$