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Math Help - excercise on the newton's binomial

  1. #1
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    excercise on the newton's binomial

    forgive me for my horrible english (I'm italian and I'have registred for this forum to test my language skills and to better understand the maths).
    This is my question: how can i prove with the Newton's binomial that (1+1)^n=2^n. Obviously is easy adding 1 with 1, but the book where i've found this exercise (it's "what's the mathematics?", Courant and Robbins, it's very famous) explicitly ask to use the binomial's theorem.
    Thanks for the attention and for the patience.

    P.s: How can I write the binomial's Coefficients on the forum?
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  2. #2
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    If you use the binomial theorem with a and b equal to 1 you are left with

    \left(\begin{array}{c}n\\0\end{array}\right) +\left(\begin{array}{c}n\\1\end{array}\right) +........+\left(\begin{array}{c}n\\n\end{array}\ri  ght)

    Which is equal to 2^{n}

    You can prove it with induction, committee argument, etc.
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  3. #3
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    Quote Originally Posted by pippo93 View Post
    [snip]
    How can I write the binomial's Coefficients on the forum?
    Here's the code galactus used to get \left(\begin{array}{c}n\\1\end{array}\right):

    [tex]\left(\begin{array}{c}n\\1\end{array}\right)[/tex]

    Clever.

    But another (easier way) way to get {n \choose 1} is:

    [tex]{n \choose 1}[/tex]
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  4. #4
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    Thank you Mr. Fantastic. I knew there was an easier way, but I couldn't remember it.
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  5. #5
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    BTW, Mr. Fantastic, I am going to return to consume your puny planet.
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  6. #6
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    Quote Originally Posted by galactus View Post
    BTW, Mr. Fantastic, I am going to return to consume your puny planet.
    Give it up, big guy. How many times have you tried and failed!

    (I was wondering when we'd finally 'meet' )
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  7. #7
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    Quote Originally Posted by galactus View Post
    If you use the binomial theorem with a and b equal to 1 you are left with

    \left(\begin{array}{c}n\\0\end{array}\right) +\left(\begin{array}{c}n\\1\end{array}\right) +........+\left(\begin{array}{c}n\\n\end{array}\ri  ght)

    Which is equal to 2^{n}

    You can prove it with induction, committee argument, etc.

    How do you prove it with induction?
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  8. #8
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    Quote Originally Posted by pippo93 View Post
    How do you prove it with induction?
    Where are you stuck with the proof by induction?
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    Where are you stuck with the proof by induction?
    i'm not sure that my proof are correct and that in my proof i use the newton's binomial.

    if I want prove it with induction I verify basis case with n=0 and n=1 it's banal. Then I prove that if (1+1)^n=2^n is true it must be true also for n+1. (1+1)^n +^1=(1+1)^n(1+1) is true (if is true (1+1)^n=2^n)
    Than it's proved? But i didn't use the binomila's theorem
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