Question

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

• A. $3^{2010}$
• B. $1$
• C. $2^{2010}$
• D. $4^{2010}$

Answer 1

The correct option is C $2^{2010}$

We know that, $1,wand{w}^2$ are three cube roots of unity,

And, $(1+w+w^2=0)$ and $w^3=1$

Put $x=w$, we get

$(3+w+w^2{)}^{2010}=a_0+a_{1}w+a_2{w}^2+a_3{\omega }^3+\cdots$

$(2+1+w+w^2{)}^{2010}=a_0+a_{1}w+a_2{w}^2+a_3{\omega }^3+\cdots$
$\Rightarrow 2^{2010}=a_0+a_{1}w+a_2{w}^2+a_3+\cdots \dots \left(1\right)$
Put $x=w^2$,we get
$(3+w^2+w^4{)}^{2010}=a_0+a_1{w}^2+a_2{w}^4+a_3{\omega }^6+\cdots$
$\Rightarrow 2^{2010}=a_0+a_1{w}^2+a_{2}w++a_3+\cdots \dots \left(2\right)$
Adding $\left(1\right),\left(2\right)$, we get
$\Rightarrow 2\times 2^{2010}=2a_0-a_1-a_2+2a_3-a_4-a_5+2a_6+\cdots$
$\Rightarrow 2^{2010}=a_0-\frac{1}{2}{a}_1-\frac{1}{2}{a}_2+a_3-\frac{1}{2}{a}_4-\frac{1}{2}{a}_5+a_6\cdots$