Question

If $(3+x^{2008}+x^{2009}{)}^{2010}=a_0+a_{1}x+a_2{x}^2+...+a_n{x}^n,$ then the value of $a_0-\frac{1}{2}{a}_1-\frac{1}{2}{a}_2+a_3-\frac{1}{2}{a}_4-\frac{1}{2}{a}_5+a_6-...$ is

• A. $3^{2010}$
• B. $1$
• C. $2^{2010}$
• D. none of these

Answer 1

Answer: (C)

Put $x=\omega ,{\omega }^2$
${\left(3+\omega +{\omega }^2\right)}^{2010}=a_0+a_1\omega +a_2{\omega }^2+...$
$\Rightarrow 2^{2010}=a_o+a_1{\omega }^2+a_2\omega +a_3+a_4\omega +.....\left(1\right)$
and $2^{2010}=a_0+a_1{\omega }^2+a_2\omega +a_3+a_4\omega +.....\left(2\right)$
Adding $\left(1\right)and\left(2\right),$ we have
$2\times 2^{2010}=2a_o-a_1-a_2+2a_3-a_4-a_5+2a_6-...$
or $2^{2010}=a_o-\frac{1}{2}{a}_1-\frac{1}{2}{a}_1+a_3-\frac{1}{2}{a}_4-\frac{1}{2}{a}_5+a_6...$