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Math Help - Binomial Proof

  1. #1
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    Post Binomial Proof

    I havn't done this in a long time! And apparently I should know this easy, it sort of looks like a proof by induction, which I havn't done before and I am frantically trying to learn!

    Show that for each integer n the alternating sum of binomial coefficients:

    1 - (n) + ... + (-1)^k(n) + ... + (-1)^(n-1)( n ) + (-1)^n
    .....(1)....................(k)................... .......(n-1)
    is zero. What is the value of the sum

    (so what ive done here is started with an "inductive basis of n=1 which kind of suggests it goes to zero but without the appropriate conciseness)

    1 + (n) + ... + (n) + ... + ( n ) +1
    ......(1)...........(k)...........(n-1)

    I understand the layout is a bit rubbish but I hope you can fathom it!
    Any help would be greatly appreciated!
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  2. #2
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    Quote Originally Posted by a8281333 View Post
    I havn't done this in a long time! And apparently I should know this easy, it sort of looks like a proof by induction, which I havn't done before and I am frantically trying to learn!

    Show that for each integer n the alternating sum of binomial coefficients:

    1 - (n) + ... + (-1)^k(n) + ... + (-1)^(n-1)( n ) + (-1)^n
    .....(1)....................(k)................... .......(n-1)
    is zero. What is the value of the sum

    (so what ive done here is started with an "inductive basis of n=1 which kind of suggests it goes to zero but without the appropriate conciseness)

    1 + (n) + ... + (n) + ... + ( n ) +1
    ......(1)...........(k)...........(n-1)

    I understand the layout is a bit rubbish but I hope you can fathom it!
    Any help would be greatly appreciated!

    For any two real (or even complex) numbers a,b:\,\,(a+b)^n=\sum\limits^n_{k=0}\binom{n}{k}a^{  n-k}b^k\,,\,\,n\in\mathbb{N} .

    You have \sum\limits^n_{k=0}\binom{n}{k}(-1)^k=\sum\limits^n_{k=0}\binom{n}{k}1^{n-k}(-1)^k=... and the result is immediate.

    Tonio
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  3. #3
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    so basically, as k is even then odd, they alternate, basically giving a sequence that cancels out? I hope this is what you mean, I'm really keen to learn the process in detail to build a firm fundamental understanding
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  4. #4
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    Quote Originally Posted by a8281333 View Post
    so basically, as k is even then odd, they alternate, basically giving a sequence that cancels out? I hope this is what you mean, I'm really keen to learn the process in detail to build a firm fundamental understanding

    No, I did not mean that, though that is what happens. Do read again CAREFULLY and pay attention to the (a+b)^n=... thing

    Tonio
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  5. #5
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    Sorry, I dont understand, this must be really frustrating for you! I best get my head in a book, I havn't done this sort of mathematics in two years so I'm very rusty, but I'm sure once I discover the answer/steps it will all come rushing back!
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  6. #6
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    after a bit of research, am i right to assume this is a telescoping series?
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  7. #7
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    Update, apparently I dont need induction? I just expand (1+1)^n and (1-1)^n, that doesnt seem right to me though? anyone care to shread any light on this?
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  8. #8
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    Quote Originally Posted by a8281333 View Post
    Update, apparently I dont need induction? I just expand (1+1)^n and (1-1)^n, that doesnt seem right to me though? anyone care to shread any light on this?
    More like just the second one. I think you didn't pay attention to what Tonio was saying.

    The binomial theorem states \displaystyle \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k = (a+b)^n

    What the sum you have is \displaystyle \sum_{k=0}^{n}\binom{n}{k}(1)^{n-k}(-1)^k

    Don't you see what a and b are in this case?
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  9. #9
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    Ah yes! I see, it is just (1-1)^n, then the second part is just 2^n!, how could i be so stupid! I love learning so my solution for part 2 is just (1+1)^n which is just 2^n right?! Woop
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  10. #10
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    Quote Originally Posted by a8281333 View Post
    so my solution for part 2 is just (1+1)^n which is just 2^n right?! Woop
    Yes.
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