Question

If ${(x^{2006}+\frac{1}{x^{2006}}+2)}^{2010}=a_0+a_{1}x+a_2{x}^2+....a_n{x}^n$, then the value of $2a_0-a_1-a_2+2a_3-a_4-a_5+2a_6-...$ is

• A. $2$
• B. $1$
• C. $0$
• D. $3$

Answer 1

The correct option is A $2$

$2a_0-a_1-a_2+2a_3-a_4-a_5+2a_6-...$
Here, $1$st, $4$th, $7$th and so on terms have same coefficient.
$2$nd, $5$th, $8$th and so on terms have same coefficient.
$3$rd, $6$th, $9$th and so on terms have same coefficient.

If $x=\omega$, where $\omega$ is cube root of unity
${({\omega }^{2006}+\frac{1}{{\omega }^{2006}}+2)}^{2010}=a_0+a_1\omega +a_2{\omega }^2+...+a_n{\omega }^n$
$\Rightarrow ({\omega }^2+\omega +1+1{)}^{2010}=a_0+a_1\omega +a_2{\omega }^2+...+a_n{\omega }^n$
Since, $1+\omega +{\omega }^2=0$
$\Rightarrow 1=a_0+a_1\omega +a_2{\omega }^2+...+a_n{\omega }^n...\left(1\right)$

If $x={\omega }^2$
${({\omega }^{4012}+\frac{1}{{\omega }^{4012}}+2)}^{2010}=a_0+a_1{\omega }^2+a_2{\omega }^4+...+a_n{\omega }^{2n}$
$\Rightarrow (\omega +{\omega }^2+1+1{)}^{2010}=a_0+a_1{\omega }^2+a_2{\omega }^4+...+a_n{\omega }^{2n}$
$\Rightarrow 1=a_0+a_1{\omega }^2+a_2{\omega }^4+...+a_n{\omega }^{2n}....\left(2\right)$

$\left(1\right)+\left(2\right)$, we get
$2=2a_0+a_1(\omega +{\omega }^2)+a_2({\omega }^2+{\omega }^4)+a_3({\omega }^3+{\omega }^6)+...+a_n({\omega }^n+{\omega }^{2n})$
$\Rightarrow 2=2a_0-a_1-a_2+2a_3-a_4-a_5+2a_6-...$

$\therefore$ solution is $2$