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Math Help - Binomial Theorem - Proof

  1. #1
    Member mybrohshi5's Avatar
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    Binomial Theorem - Proof

    Show that, for n > 0,

     \sum_{i=0}^{n}(-1)^i\binom{n}{i} = 0

    I know i am supposed to use the binomial theorem which is just,

     (a+b)^n = \sum_{i=0}^{n}\binom{n}{i} a^i b^{n-i}

    I am really just not sure where to start though

    Thank you for any help/tips

    (if this needs to be moved to Discrete Mathematics, Set Theory and Logic, would a mod please move it. thank you)
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  2. #2
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    Hint: \displaystyle \sum_{i=0}^{n}\binom{n}{i}(-1)^i = \sum_{i=0}^{n}\binom{n}{i}(-1)^i(1)^{n-i}.
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  3. #3
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    Here is a different hint.
    0=(1-1)^n=~? That is of course a sum.
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  4. #4
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    Or, you could let

    a=b=-\frac{1}{2}

    or something similar and plug them into the expansion, noticing a pattern.
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  5. #5
    Member mybrohshi5's Avatar
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    Ok great i think i got it.

    for a = 1 and b = -1 and using any number where n > 0 you can see that when plugged into the binomial theorem it is a repeating pattern of (in this case) 2 + -2 + 2 + -2 .... which therefore will = 0 proving \sum_{i=0}^{n}(-1)^i\binom{n}{i} = 0 to be true.

    Does this seem correct or am i missing something? I am very new to proofs (this is actually my first one haha )
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  6. #6
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    Quote Originally Posted by mybrohshi5 View Post
    Ok great i think i got it.

    for a = 1 and b = -1 and using any number where n > 0 you can see that when plugged into the binomial theorem it is a repeating pattern of (in this case) 2 + -2 + 2 + -2 .... which therefore will = 0 proving \sum_{i=0}^{n}(-1)^i\binom{n}{i} = 0 to be true.

    Does this seem correct or am i missing something? I am very new to proofs (this is actually my first one haha )
    Why would you get 2+(-2)+2+(-2)+.... ?

    What would the \binom{n}{i} values be ?

    and "yes", you are missing something, not only that n could be even giving an odd number of terms,
    so the sum would not be zero.

    TheCoffeeMachine and Plato are drawing your attention to the fact that

    \displaystyle\ (1-1)^n=\binom{n}{0}(-1)^01^n+\binom{n}{1}(-1)^11^{n-1}+\binom{n}{2}(-1)^21^{n-2}+.....+\binom{n}{n-1}(-1)^{n-1}1^1+\binom{n}{n}(-1)^n1^0

    The expansion is \displaystyle\sum_{i=0}^n\binom{n}{i}(-1)^i1^{n-i}

    and 1-1=0
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  7. #7
    Member mybrohshi5's Avatar
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    I have been looking over these posts again for the past day or so and I still can't seem to wrap my mind around this.

    I get using A = -1 and B = 1 and using the binomial theorem will give you the expansion

     \displaystyle\ (-1 + 1)^n=\binom{n}{0}(-1)^01^n+\binom{n}{1}(-1)^11^{n-1}+\binom{n}{2}(-1)^21^{n-2}+.....+\binom{n}{n-1}(-1)^{n-1}1^1+\binom{n}{n}(-1)^n1^0

    I guess i just dont get how this proves that for n > 0

     \displaystyle \sum_{i=0}^{n}\binom{n}{i}(-1)^i = 0

    I feel like i am making this much more difficult than it should be
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  8. #8
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    It's because

    1^{n-i}=1

    in each term of the expansion.

    TheCoffeeMachine related the binomial expansion to your closed-form sum.

    They are the same.
    Hence there is no need to calculate each term of the expansion, since the sum is

    (1-1)^n=0^n=0

    as Plato also showed.
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  9. #9
    Member mybrohshi5's Avatar
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    Thank you for the great explanation I am going to give my brain a rest then rework the proof and make sure i understand everything and if I dont I will post back again

    Thanks again for all the replies everyone. I really appreciate the help
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  10. #10
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    You said you are supposed to use the binomial theorem. So for n > 0 we have,
    (a+b)^n = \sum_{i=0}^{n}\binom{n}{i} a^i b^{n-i}
    let a = -1 and b = 1. This equality then becomes,
    ((-1)+(1))^n = \sum_{i=0}^{n}\binom{n}{i} (-1)^i (1)^{n-i}
    re-written,
    (0)^n = \sum_{i=0}^{n}\binom{n}{i} (-1)^i

    So the binomial theorem actually proves it for you. You just need to plug in those values. This is just a special case of the binomial theorem.

    You could also use the binomial theorem to show that
    ((1)+(1))^n = \sum_{i=0}^{n}\binom{n}{i} (1)^i (1)^{n-i}
    So that
    (2)^n = \sum_{i=0}^{n}\binom{n}{i}

    in each case you are using the special cases of a theorem that has already been proven for you. I hope that helps you out a bit.
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