Question

$(1+x{)}^n=P_0+P_{1}x+P_2{x}^2+P_n{x}^n$, prove that :
$P_3-P_7+P_{11}+....=\frac{1}{2}(2^{n-1}-2^{\frac{n}{2}}sin\frac{nx}{4})$

Answer 1

$(1+x{)}^n=p_0+p_{1}x+p_2{x}^2+p_3{x}^3+.....$
Multiplying both side by x,
$x(1+x{)}^n=p_{0}x+p_1{x}^2+p_2{x}^3+p_3{x}^4+.....$
If $x=1^{1/4}then{x}^4-1=0$
or $(x^2-1)(x^2+1)=0$
$\therefore x=1,-1,i,-iand{x}^4=1$
Also 1 + 1 + i - i = 0
Put x = 1, -1, i, -i in both sides of (1) and add
$2^n+0+i(1+i{)}^n-i(1-i{)}^n$
$=4(p_3+p_7+p_{11}+.....)$
$1+i=\sqrt{2}{e}^{\pi i/4},1-i=\sqrt{2}{e}^{-\pi i/4}$
L.H.S. = $2^n+i.2^{n/2}[e^{n\pi i/4}-e^{-n\pi i/4}]$
$=2^n-i{.2}^{n/2}2isin\frac{n\pi }{4}$
$=2^n-{2.2}^{n/2}sin\frac{n\pi }{4}=4E$
$\therefore E=\frac{1}{2}\{2^{n-1}-2^{n/2}sin\frac{n\pi }{4}\}$