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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1
Let [x]\left[ x \right] denote greatest integer less than or equal to xx. If for n∈N, (1−x+x3)n=∑j=03najxjn \in N,\,{\left( {1 - x + {x^3}} \right)^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}}, then ∑j=0[3n2]a2j+4∑j=0[3n−12]a2j+1\sum\limits_{j = 0}^{\left[ {\frac{{3n}}{2}} \right]} {{a_{2j}} + 4} \sum\limits_{j = 0}^{\left[ {\frac{{3n - 1}}{2}} \right]} {{a_{2j + 1}}} is equal to:
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Correct option: A
1
Option A (1) is correct because by substituting x=1 and x=-1 into the given equation and adding/subtracting, we can isolate the required summation and find its value.

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