Let [ x ] denote greatest integer less than or equal to x. If for n\inN, {(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 :

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Answer: C. 2 {(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}} (1 - x + {x^3}) = {a_0} + {a_1}x + {a_2}{x^2} + ...... + {a_{3n}}{x^{3n}} Put x = 1 1 = {a_0} + {a_1} + {a_2} + {a_3} + {a_4} + ........ + {a_{3n}} ...... (1) Put x = -1 1 = {a_0} - {a_1} + {a_2} - {a_3} + {a_4} + ........( - 1){}^{3n}{a_{3n}} ..... (2) Add (1) + (2) \Rightarrow {a_0} + {a_2} + {a_4} + {a_6} + ...... = 1 Sub (1) - (2) \Rightarrow {a_1} + {a_3} + {a_5} + {a_7} + ...... = 0 Now, \sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} ight]} {{a_{2j}}} + 4\sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} ight]} {{a_{2j }}} + 1 = ({a_0} + {a_2} + {a_4} + ......) + 4({a_1} + {a_3} + .....) = 1 + 4 	imes 0 + 1 = 1 + 1 = 2

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