Question

If for $z$ as real or complex, $(1+z^2+z^4{)}^8=C_0+C_1{z}^2+C_2{z}^4+\cdots +C_{16}{z}^{32}$, then

• A. $C_0-C_1+C_2-C_3+\dots +C_{16}=1$
• B. $C_0+C_3+C_6+C_9+C_{12}+C_{15}=3^7$
• C. $C_2+C_5+C_8+C_{11}+C_{14}=3^6$
• D. $C_1+C_4+C_7+C_{10}+C_{13}+C_{16}=3^7$

Answer 1

Answer: (A), (B), (D)

$C_1+C_4+C_7+C_{10}+C_{13}+C_{16}=3^7$

$(1+z^2+z^4{)}^8=C_0+C_1{z}^2+C_2{z}^4+\dots +C_{16}{z}^{32}\dots [1]$

Putting $z=i$, where $i=\sqrt{-1}$,
$(1-1+1{)}^8=C_0-C_1+C_2-C_3+\dots +C_{16}$
or, $C_0-C_1+C_2-C_3+\dots +C_{16}=1$

Also, putting $z=\omega$, where $\omega$ is cube root of unity
$(1+{\omega }^2+{\omega }^4{)}^8=C_0+C_1{\omega }^2+C_2{\omega }^4+\dots +C_{16}{\omega }^{32}$
$\Rightarrow C_0+C_1{\omega }^2+C_2\omega +\dots +C_{16}{\omega }^2=0\dots \left[2\right]$

Putting $z={\omega }^2$,
$(1+{\omega }^4+{\omega }^8{)}^8=C_0+C_1{\omega }^4+C_2{\omega }^8+\dots +C_{16}{\omega }^{64}$
$\Rightarrow C_0+C_1\omega +C_2{\omega }^2+\dots +C_{16}\omega =0\dots \left[3\right]$

Putting $z=1$,
$3^8=C_0+C_1+C_2+\dots +C_{16}\dots \left[4\right]$

Adding $\left[2\right],\left[3\right],\left[4\right]$, we get
$3(C_0+C_3+C_6+\dots +C_{15})=3^8$
$\Rightarrow C_0+C_3+C_6+\dots +C_{15}=3^7$

Similarly, first multiplying $\left[1\right]$ by $z$ and then putting $1,\omega ,{\omega }^2$ and adding, we get
$C_1+C_4+C_7+C_{10}+C_{13}+C_{16}=3^7$

Multiplying $\left[1\right]$ by $z^2$ and then putting $1,\omega ,{\omega }^2$ and adding, we get
$C_2+C_5+C_8+C_{11}+C_{14}=3^7$