Question

Let $\left[x\right]$ denote the greatest integer less than or equal to x. If for $n\in N,(1–x+x^3)n=\sum _{j=0}^{3n}{a}_j{x}^j,then\sum _{j=0}^{\left(\frac{3n}{2}\right)}{a}_{2j}+4\sum _{j=0}^{\left(\frac{3n-1}{2}\right)}{a}_{2j}+1=$

• A. $1$
• B. $n$
• C. $2^n-1$
• D. 2

Answer 1

The correct option is A

$1$

It is given that:

$\begin{array}{rcl}(1–x+x^3)n& =& \sum _{j=0}^{3n}{a}_j{x}^j\\ & =& a_0+a_{1}x+a_2{x}^2+.......a_{3n}{x}^{3n}\\ & & \end{array}$

Put $x=1$, we get:

$\begin{array}{rcl}1& =& a_0+a_1+a_2....+a_{3n}\left(1\right)\end{array}$

On putting $x=-1$, we get:

$\begin{array}{rcl}1& =& a_0-a_1+a_2....{(-1)}^{3n}{a}_{3n}\left(2\right)\end{array}$

Now, $\left(1\right)+\left(2\right)$

$\begin{array}{rcl}{a}_0+a_2+a_4+a_6.....& =& 1\end{array}$

And, $\left(1\right)-\left(2\right)$

$\begin{array}{rcl}{a}_1+a_3+a_5+a_7.....& =& 0\end{array}$

On calculating:

$$\begin{gathered} \sum _{j=0}^{\left(\frac{3n}{2}\right)}{a}_{2j}+4\sum _{j=0}^{\left(\frac{3n-1}{2}\right)}{a}_{2j}+1 \\ =\left(a_0+a_2+a_4....\right)+4\left(a_0+a_1+a_3....\right) \\ =1+4\left(0\right) \\ =1 \end{gathered}$$

Hence, option A is the correct answer.