Question:
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 \(\sum\limits_{j=0}^{\left[\frac{3n}{2}\right]} a_{2j} + 4 \sum\limits_{j=0}^{\left[\frac{3n-1}{2}\right]} a_{2j+1}\) is equal to :
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 \(\sum\limits_{j=0}^{\left[\frac{3n}{2}\right]} a_{2j} + 4 \sum\limits_{j=0}^{\left[\frac{3n-1}{2}\right]} a_{2j+1}\) is equal to :
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Substitute \(x=1\) and \(x=-1\) to get the sum of even-indexed and odd-indexed coefficients.
Updated On: Nov 4, 2026
- \(2^{n-1}\)
- \(n\)
- \(2\)
- \(1\)
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The Correct Option is D
Solution and Explanation
Let \(E=\sum a_{2j}\) (even-indexed coefficients) and \(O=\sum a_{2j+1}\) (odd-indexed). The upper limits \([3n/2]\) and \([(3n-1)/2]\) cover all even and odd indices up to \(3n\).
\(x=1\): \(E+O = (1-1+1)^n = 1\).
\(x=-1\): \(E-O = (1+1-1)^n = 1\).
So \(E = 1,\ O = 0\). The required value is \(E + 4O = 1 + 0 = 1\).
Check n = 1: \(1 - x + x^3\): E = a0 + a2 = 1, O = a1 + a3 = -1 + 1 = 0.
Answer: (4) \(1\)
\(x=1\): \(E+O = (1-1+1)^n = 1\).
\(x=-1\): \(E-O = (1+1-1)^n = 1\).
So \(E = 1,\ O = 0\). The required value is \(E + 4O = 1 + 0 = 1\).
Check n = 1: \(1 - x + x^3\): E = a0 + a2 = 1, O = a1 + a3 = -1 + 1 = 0.
Answer: (4) \(1\)
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