Let [x] denote greatest integer less than or equal to x. If for n ∈ N,
Let [x] denote greatest integer less than or equal to x. If for n ∈ \(\mathbb N\) \((1-x+x^3)^n\) = \(\displaystyle\sum_{j=0}^{3n} a_jx^j\) then \(\displaystyle\sum_{j=0}^{\big[\frac{3n}{2}\big]} a_{2j}+4\displaystyle\sum_{j=0}^{\big[\frac{3n-1}{2}\big]} a_{2j}+1\) is equal to :(1) 2(2) \(2^{n-1}\)(3) 1(4) n
Answers
The sum in the question can be rearranged as: \(\displaystyle\sum_{j=0}^{\big[\frac{3n}{2}\big]} a_{2j}+4\displaystyle\sum_{j=0}^{\big[\frac{3n-1}{2}\big]} a_{2j+1}\) Note that since the polynomial is symmetric, \(a_{2j+1}=a_{2j}\) for all \( j\in \mathbb N\). Therefore, the sum simplifies to \(2 \displaystyle\sum_{j=0}^{\big[\frac{3n}{2}\big]} a_{2j} + 1 = 2a_0 + 1\) Now, since \((1-x+x^3)^n\) = \(\displaystyle\sum_{j=0}^{3n} a_jx^j\), substituting \(x=0\) gives \(1 = a_0\) Therefore, the sum is equal to 1. So, the answer is (3).
