Question

If $C_0,C_1,C_2,....C_n$ are binomial coefficient in the expansion of $(1+x{)}^n$, then value of $C_1+C_4+C_7+...$ equals

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

Answer 1

The correct option is B $\frac{1}{3}(2^n-\cos \frac{n\pi }{3}+\sqrt{3}\sin \frac{n\pi }{3})$

Given ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+\cdots +C_n{x}^n$
$\therefore x^2{(1+x)}^n=x^2{C}_0+C_1{x}^3+C_2{x}^4+\cdots +C_n{x}^{n+2}(\ast )$
Now substituting $x=1$,$\omega ,{\omega }^{2}in(\ast ),$ we get $2^n=C_0+C_1+C_2+C_3+\cdots +C_n\cdots \left(1\right)$
${\omega }^2{(1+\omega )}^n=C_0{\omega }^2+C_1{\omega }^3+C_2{\omega }^4+\cdots +C_n.{\omega }^{n+2}$...(2)
${\omega }^4{(1+{\omega }^2)}^n=C_0{\omega }^4+C_1{\omega }^6+C_2{\omega }^8+\cdots +C_n.{\omega }^{2n+4}$...(3)
Now adding (1), (2) and (3) we get $2^n+\omega {,}^2{(1+\omega )}^n+{\omega }^4{(1+{\omega }^2)}^n$
$=C_0(1+\omega +{\omega }^2)+C_1(1+{\omega }^3+{\omega }^6)+C_2(1+\omega +{\omega }^2)$
$+....=3(C_1+C_4+C_7+\cdots )$ (other terms vanished).$3(C_1+C_4+C_7+\cdots )$ $=2^n+{\omega }^2(\cos \frac{n\pi }{2}+i\sin \frac{n\pi }{3})+{\omega }^4(\cos \frac{n\pi }{3}-i\sin \frac{n\pi }{3})$ $=2^n+({\omega }^2+\omega )\cos \frac{n\pi }{3}+i({\omega }^2-\omega )\sin \frac{n\pi }{3}$ $=2^n-\cos \frac{n\pi }{3}+i(-\sqrt{3i})\sin \frac{n\pi }{3}$ As $\omega =\frac{-1+i\sqrt{3}}{2}and{\omega }^2=\frac{-1-i\sqrt{3}}{2}$ ${\omega }^2-\omega =-\frac{1}{2}-\frac{i\sqrt{3}}{2}+\frac{1}{2}-\frac{i\sqrt{3}}{2}=i\sqrt{3}$ ${\omega }^2-\omega =-\sqrt{3}$ $\therefore 3(C_1+C_4+C_7\cdots )=2^n-\cos \frac{n\pi }{3}+\sqrt{3}\sin \frac{n\pi }{3}$ $C_1+C_4+C_7+\cdots =\frac{1}{3}(2^n-\cos \frac{n\pi }{3}+\sqrt{3}\sin \frac{n\pi }{3})$