Question

If $n$ is a multiple of $3,$ then $C_0+C_3+C_6+\dots .$ is equal to

• A. $\frac{2^n+2}{3}$
• B. $\frac{2^n-2}{3}$
• C. $\frac{2^n+2(-1{)}^{n/3}}{3}$
• D. $\frac{2^n-2(-1{)}^n}{3}$

Answer 1

The correct option is A $\frac{2^n+2}{3}$

Question - if $n$ is a multiple of 3 , the $C_0+C_3+C_6\dots$ is equal to

(A) $\frac{2^n+2}{3}$

(c) $\frac{2^n+2(-1{)}^{n/3}}{3}$

(B) $\frac{2^n-2}{3}$

(D) $\frac{2^n-2(-1{)}^n}{3}$

Solution $-(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\dots +\mathrm{Cn}{x}^n$

Substitute $x=1$ in

(1) $\Rightarrow 2^n=C_0+C_1+C_2+C_3+\dots .+C_n-$

Substitute $x=\omega$ in

(1)

$\Rightarrow (1+\omega {)}^n=C_0+C_1\omega +C_2{\omega }^2+\dots .+C_n{\omega }^n$

We know that $1+\omega +{\omega }^2=0$ and ${\omega }^3=1$

$\Rightarrow {(-{\omega }^2)}^n=C_0+C_1\omega +C_2{\omega }^2+C_3+\dots .+C_n{\omega }^n$

$\text{Substitute}x={\omega }^2\mathrm{in}\left(1\right)$

$\Rightarrow {(1+{\omega }^2)}^n=C_0+C_1{\omega }^2+C_2{\omega }^4+\dots +C_n{\omega }^{2n}$

$\Rightarrow (-\omega {)}^2=C_0+C_1{\omega }^2+C_2\omega +\dots +C_n{\omega }^{2n}-(4)$

(2)+(3)+(4)

$\Rightarrow 2^n+(-1{)}^n({\omega }^{2n}+{\omega }^n)=[3C_0+C_1(1+\omega +{\omega }^2)+C_2(1+\omega +{\omega }^2)+$

$C_3+\dots ]$

$\Rightarrow 3(C_0+C_3+C_6+\dots )=2^n+(-1{)}^n({\omega }^{2n}+{\omega }^n)$

${\omega }^{2n}+{\omega }^n=e^{2inJ}3+e^{4i\frac{nJ}{4}}=(\cos \frac{2n\pi }{3}+\cos \frac{4n\pi }{3})+i(\sin \frac{2n\pi }{3}+\sin \frac{4n\pi }{3})$

$=2\cos \left(n\pi \right)\cos (-\frac{n\pi }{3})+2i\sin \left(n\pi \right)\cos (-\frac{n\pi }{3})$

$=2(-1{)}^n\cos \left(\frac{n\pi }{3}\right)$

$\Rightarrow C_0+C_3+C_6+\dots =\frac{2^n+2\cos \frac{n\pi }{3}}{3}$

$\Rightarrow C_0+C_3+C_6+\dots .=\frac{2^n+2}{3}$

$\text{As given that}n\text{is a multiple of}3$

Hence, (A) is the correct option.