MathematicsMedium64×since 2002Q2993Let [ x ] denote greatest integer less than or equal to x. If for n∈\in∈N, (1−x+x3)n=∑j=03najxj{(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}}(1−x+x3)n=j=0∑3najxj, then ∑j=0[3n2]a2j+4∑j=0[3n−12]a2j+1\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}j=0∑[23n]a2j+4j=0∑[23n−1]a2j+1 is equal to :A2^n −-− 1BnC2D1Check answerSkip