Question

Let
$(1+x+x^2{)}^{2014}=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+....+a_{4028}{x}^{4028}$
and let
$A=a_0-a_3+a_6-.....+a_{4026}$,
$B=a_1-a_2+a_7-.....+a_{4027}$,
$C=a_2-a_5+a_8-.....+a_{4028}$
Then

• A. $|A|=|B|>|C|$
• B. $|A|=|B|<|C|$
• C. $|A|=|C|>|B|$
• D. $|A|=|C|<|B|$

Answer 1

The correct option is D $|A|=|C|<|B|$

$(1+x+x^2{)}^{2014}=a_0+a_{1}x+a_2{x}^2+...+a_{4028}{x}^{4028}$

substituting $x=-1,-\omega ,-{\omega }^2$ where $\omega =e^{i2\pi /3}$

$1=a_0-a_1+a_2-a_3+a_4-a_5+a_6...+a_{4028}$ ...........(i)

$(1-\omega +{\omega }^2{)}^{2014}=a_0-a_1\omega +a_2{\omega }^2-a_3+a_4\omega -a_5{\omega }^2+a_6...+a_{4028}{\omega }^2$ .........(ii)

$(1-{\omega }^2+{\omega }^4{)}^{2014}=a_0-a_1{\omega }^2+a_2\omega -a_3+a_4{\omega }^2-a_5\omega +a_6...+a_{4028}\omega$ .........(iii)

$\text{(i)}+\text{(ii)}+\text{(iii)}\Rightarrow a_0-a_3+a_6...+a_{4026}=\text{A}=\frac{1+2^{2014}(\omega +{\omega }^2)}{3}$

$\Rightarrow \text{|A|}=\frac{2^{2014}-1}{3}$

$\text{(i)}+{\omega }^2\text{(ii)}+\omega \text{(iii)}\Rightarrow a_1-a_4+a_7...+a_{4027}=\text{B}=-\frac{1+2^{2015}}{3}$

$\text{(i)}+\omega \text{(ii)}+{\omega }^2\text{(iii)}\Rightarrow a_2-a_5+a_8...+a_{4028}=\text{C}=-\frac{1-2^{2014}}{3}$

$\Rightarrow \text{|A| = |C| < |B|}$