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Math Help - equality proof

  1. #1
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    equality proof

    equality proof-delelte.jpg

    anybody knows how to solve this ? im reviewing for my test,any help is highly appreciated
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  2. #2
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    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
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  3. #3
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    i know these binomial representations! but i dont know how to solve the problem
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  4. #4
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    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})
    Last edited by veileen; March 26th 2011 at 10:41 PM.
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  5. #5
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    thank you for your reply but i really see no connection between what you wrote and the question :S:S:S
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  6. #6
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    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.
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