1. ## equality proof

anybody knows how to solve this ? im reviewing for my test,any help is highly appreciated

2. Indication:
$(1-1)^{n}=\sum_{k=0}^{n} \binom{n}{k} 1^{n-k}(-1)^{k}$
$(1+1)^{n}=\sum_{k=0}^{n} \binom{n}{k} 1^{n-k}1^{k}$
$(1+i)^{n}=\sum_{k=0}^{n} \binom{n}{k} 1^{n-k}i^{k}$
$(1-i)^{n}=\sum_{k=0}^{n} \binom{n}{k}1^{n-k}(-i)^{k}$
$i^{2}=-1$

3. i know these binomial representations! but i dont know how to solve the problem

4. Uhm, i thought it can be easily seen... ^^'
$(1-1)^{n}=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\binom{n}{4}-\binom{n}{5}+...$
$(1+1)^{n}=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\ binom{n}{3}+\binom{n}{4}+\binom{n}{5}+...$
$(1-i)^{n}=\binom{n}{0}-i\binom{n}{1}-\binom{n}{2}+i\binom{n}{3}+\binom{n}{4}-i\binom{n}{5}+...$
$(1+1)^{n}=\binom{n}{0}+i\binom{n}{1}-\binom{n}{2}-i\binom{n}{3}+\binom{n}{4}+i\binom{n}{5}+...$
$4S=(1+1)^{n}+(1-1)^{n}+(1+i)^{n}+(1-i)^{n}$
You also know that $1+i=\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})$ and $1-i=\sqrt{2}(\cos \frac{\pi}{4}-i\sin \frac{\pi}{4})$

5. thank you for your reply but i really see no connection between what you wrote and the question :S:S:S

6. I'm sorry, I wasn't attentive - at 1+i and 1-i trigonometric forms.
I noted with S the left side. If you take a sheet, a pencil and add those binomial representation, you will find S multiplied by four.
Then use Moivre for $(1+i)^{n}$ and $(1-i)^{n}$, you have:

$4S=(1+1)^{n}+(1-1)^{n}+(1+i)^{n}+(1-i)^{n}\Rightarrow 4S=2^{n}+0^{n}+\sqrt{2^{n}}(\cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4})+\sqrt{2^{n}}(\cos \frac{n\pi }{4}-i\sin \frac{n\pi }{4})$

I hope that from here you can handle yourself.