# Show that (n 1) - (n 3) + (n 5) - (n 7) + ... = 2^n/2 cos n pi/4, and (n 0) - (n 2) + (n 4) - (n...

## Question:

Show that {eq}(n/1) - (n/3) + (n/5) - (n/7) + ... = 2^{n/2} cos \frac{n \pi}{4} {/eq}, and {eq}(n/0) - (n/2) + (n/4) - (n/6) + ... = 2^{n/2} sin \frac{ n \pi}{4}. {/eq}

## Binomial Expansion

Consider a polynomial of just two terms.Suppose it has a power and we want to algebraically expand this power.Binomial expansion deals with such kind of expansions.

Euler's Identity

{eq}e^{ix}=cosx+isinx {/eq}

{eq}(1+i)^n=2^\frac{n}{2}[cos(\frac{\pi }{4})+isin (\frac{\pi }{4})]^n=2^\frac{n}{2}[cos(\frac{n\pi }{4})+i sin(\frac{n\pi }{4})] {/eq}

Powers Of Unity

• {eq}i=\sqrt(-1) \\ i^2=-1 \\ i^3=-i \\ i^4=1 {/eq}

We have our Binomial Expansion for

{eq}\left ( 1+x \right )^{n}=1+nx+\frac{n(n-1)x^{2}}{2!}+\frac{n(n-1)(n-2)x^{3}}{3!}+.... {/eq}

Or

{eq}\left ( 1+x \right )^{n}=nC_{0}+nC_{1}x+nC_{2}x^2+nC_{3}x^3+.....+nC_{n}x^n {/eq}

Put x=i in the above equation we get

{eq}\left ( 1+x \right )^{n}=nC_{0}+nC_{1}x+nC_{2}x^2+nC_{3}x^3+.....+nC_{n}x^n \\ \left ( 1+i \right )^{n}=nC_{0}+nC_{1}i+nC_{2}i^2+nC_{3}i^3+.....+nC_{n}i^n \\ \left ( 1+i \right )^{n}=nC_{0}+nC_{1}i-nC_{2}-nC_{3}i+.....+nC_{n}i^n \\ {/eq}

Equate the imaginary parts of both sides of equation

{eq}Im(1+i)^n=i(nC_{1}-nC_{3}+-nC_{5}....) \\ 2^\frac{n}{2}sin(\frac{n\pi }{4})=nC_{1}-nC_{3}+-nC_{5}.... \\ {/eq}

Thus

{eq}nC_{1}-nC_{3}+nC_{5}-nC_{3}....=2^\frac{n}{2}sin(\frac{n\pi }{4}) {/eq}

Similarly

{eq}\left ( 1+x \right )^{n}=nC_{0}+nC_{1}x+nC_{2}x^2+nC_{3}x^3+.....+nC_{n}x^n \\ \left ( 1+i \right )^{n}=nC_{0}+nC_{1}i+nC_{2}i^2+nC_{3}i^3+.....+nC_{n}i^n \\ \left ( 1+i \right )^{n}=nC_{0}+nC_{1}i-nC_{2}-nC_{3}i+.....+nC_{n}i^n \\ {/eq}

Equate the real parts of both sides of equation

{eq}\\ Re(1+i)^n=nC_{0}-nC_{2}+nC_{}4 -... \\ Re(1+i)^n=nC_{0}-nC_{2}+nC_{4} ... \\ 2^\frac{n}{2}cos(\frac{n\pi }{4})=nC_{0}-nC_{2}+nC_{4}.. {/eq}

Thus

{eq}nC_{0}-nC_{2}+nC_{4}=2^\frac{n}{2}cos(\frac{n\pi }{4}) {/eq}