Question

If $z$ is non zero complex number and $(1+z^2+z^4{)}^8=C_0+C_1{z}^2+C_2{z}^4+...+C_{16}{z}^{32}$, then

• A. $C_0-C_1+C_2-C_3+.....+C_{16}=1$
• 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)

The correct options are
A $C_0-C_1+C_2-C_3+.....+C_{16}=1$
B $C_0+C_3+C_6+C_9+C_{12}+C_{15}=3^7$
D $C_1+C_4+C_7+C_{10}+C_{13}+C_{16}=3^7$

$(1+z^2+z^4{)}^8=C_0+C_1{z}^2+C_2{z}^4+\cdots +C_{16}{z}^{32}\cdots (1)$
Putting $z=i$
$(1{)}^8=C_0-C_1+C_2-C_3+C_4-\cdots +C_{16}$
$\Rightarrow C_0-C_1+C_2-C_3+C_4-\cdots +C_{16}=1$

Putting $z=w$,
$(1+w^2+w^4{)}^8=C_0+C_1{w}^2+C_2{w}^4+\cdots +C_{16}{w}^{32}$
$\Rightarrow C_0+C_1{w}^2+C_{2}w+C_3+\cdots +C_{16}{w}^2=0\cdots \left(2\right)$

Putting $x=w^2$,
$(1+w^4+w^8{)}^8=C_0+C_1{w}^4+C_2{w}^8+\cdots +C_{16}{w}^{64}$
$\Rightarrow C_0+C_{1}w+C_2{w}^2+\cdots +C_{16}w=0\cdots \left(3\right)$

Putting $x=1,$
$3^8=C_0+C_1+C_2+C_3+...+C_{16}\cdots \left(4\right)$

Adding $\left(2\right),\left(3\right),\left(4\right)$, we have
$\Rightarrow C_0+C_3+C_6+...+C_{15}=3^7[\because 1+\omega +{\omega }^2=0]$

Similarly, first multiplying $\left(1\right)$ by $z$ and then putting $1,w,w^2$ 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 $1,w,w^2$ and adding, we get
$C_2+C_5+C_8+C_{11}+C_{14}=3^7$