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
$A=a_0-a_3+a_6-...+a_{4026},B=a_1-a_4+a_7-...-a_{4027},C=a_2-a_5+a_8-...+a_{4028}.$
Then

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

Answer 1

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

$(1+x+x^2{)}^{2014}=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+...+a_{4028}{x}^{4028}$
Put $x=-1$
$1=a_0-a_1+a_2-a_3+a_4-a_5+a_6-.....+a_{4028}....\left(1\right)$
Put $x=-\omega$
$(1-\omega +{\omega }^2{)}^{2014}=(-2\omega {)}^{2014}=a_0-a_1\omega +a_2{\omega }^2-a_3+...+a_{4028}{\omega }^{4028}....\left(2\right)$
Put $x=-{\omega }^2$
$(1-{\omega }^2+{\omega }^4{)}^{2014}=(-2{\omega }^2{)}^{2014}=a_0-a_1{\omega }^2+a_2\omega -a_3+...+a_{4028}{\omega }^{9056}....\left(3\right)$
Add equation $\left(1\right)+\left(2\right)+\left(3\right)$
$\Rightarrow 1+(2\omega {)}^{2014}+(2{\omega }^2{)}^{2014}=3(a_0-a_3+a_6......)$
$\Rightarrow A=a_0-a_3+a_6.....=\frac{1+2^{2014}\omega +2^{2014}{\omega }^2}{3}$
$A=\frac{1-2^{2014}}{3}\Rightarrow |A|=\frac{2^{2014}-1}{3}$
Now, equation $\left(1\right)+\left(2\right)\times \omega +\left(3\right)\times {\omega }^2$
$\Rightarrow \frac{1+2^{2014}{\omega }^{2014}.\omega +2^{2014}({\omega }^2{)}^{2014}.({\omega }^2)}{3}=a_2-a_5+a_8+.......$
$\Rightarrow \frac{1+2^{2014}({\omega }^{2015}+{\omega }^{4030})}{3}=a_2-a_5+a_8+.......=C$
$C=\frac{1-2^{2014}}{3}\Rightarrow |C|=\frac{2^{2014}-1}{3}$
and similarly equation $\left(1\right)+\left(2\right)\times {\omega }^2+\left(3\right)\times \omega$
$B=\frac{1+2^{2014}.({\omega }^{2016}+{\omega }^{4029})}{3}$
$|B|=\frac{1+2^{2015}}{3}$
$\therefore |B|>|A|=|C|$