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
$2\times2^{2010}=2a_{o}-a_{1}-a_{2}+2a_{3}-a_{4}-a_{5}+2a_{6} - ...$
or $2^{2010}=a_{o}-\frac{1}{2}a_{1}-\frac{1}{2}a_{1}+a_{3}-\frac{1}{2}a_{4}-\frac{1}{2}a_{5}+a_{6}...$