Question

$(1+x{)}^n=P_0+P_{1}x+P_2{x}^2+P_n{x}^n$, prove that :
$P_1-P_4+P_7+....=\frac{1}{3}(2^n+2cos(n-2\left)\frac{\pi }{3}\right)$

Answer 1

$(1+x{)}^n=p_0+p_{1}x+p_2{x}^2+p_3{x}^3+p_4{x}^4+.....$
Keeping in view that we want $3p_1$ in (a), we multiply both side by $x^2.$
$\therefore x^2(1+x{)}^n$
$=p_0{x}^2+p_1{x}^3+p_2{x}^4+p_3{x}^5+p_4{x}^6+.....$
Now put x = 1, $\omega$, ${\omega }^2$ and add.
${1.2}^n+{\omega }^2(1+\omega {)}^n+{\omega }^4(1+{\omega }^2{)}^n$
$=3(p_1+p_4+p_7+.....)$
$\omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}=e^{2\pi /3},{\omega }^2=-\frac{1}{2}-i\frac{\sqrt{3}}{2}=e^{4\pi i/3}$
$1+\omega =\frac{1}{2}+i\frac{\sqrt{3}}{2}=e^i\pi /3$,
$1+\omega =\frac{1}{2}-i\frac{\sqrt{3}}{2}=e^{-}i\pi /3$
Hence from (1). L.H.S.
$=2^n+e^{4\pi i/3}.e^{n\pi i/3}+e^{2\pi i/3}.e^{-n\pi i/3}$
$=2^n+e^{-2\pi i/3}.e^{n\pi i/3}+e^{2\pi i/3}.e^{-n\pi i/3}$
$=2^n+e^{\frac{(n-2)\pi i}{3}}{e}^{-\frac{(n-2)\pi i}{3}}$
$2^n+2cos\frac{n-2}{3}\pi =3Setc.$