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    Can someone explain the proof of this binomial theorem:

    \displaystyle (x+y)^n=\sum_{k=0}^{n} \binom{n}{k} x^ky^{n-k}

    to me simply because I find the logic a little hard to follow along although I grasp it a small amount but would like to fully come to understand this.

    I am looking for the combinations proof not the induction.

    Thanks!
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    First of all (x + y)^n = \underbrace{(x + y)(x + y)...(x + y)}_{\text{n times}}.

    Now what you want to do to get the general formula for the binomial theorem (when n \in \mathbb{N}), is to collect all the individual x^r, \ r = 0, 1, ... n together.

    For the x^n, we see that we want n of the xs out of n, so we have \displaystyle \binom{n}{n} choices, which is equal to 1. But we have no ys in our expression, as all of our choices have been filled with xs.

    For the x^{n - 1}, we see that we need to choose n-1 xs out of n ones. Now clearly this is \displaystyle \binom{n}{n - 1}. But if you choose n - 1 xs, you always have 1 y with your x^{n - 1}, so we have that \displaystyle \binom{n}{n - 1}x^{n - 1}y.

    Same logic for x^{n - 2}. We need to choose n - 2 xs out of n, so we have \displaystyle \binom{n}{n - 2} choices. But if you wants the parts of the expression with only x^{n - 2}, then you're going to get y^2 with it, as the remaining 2 choices must be given to y. So we have that \displaystyle \binom{n}{n - 2}x^{n - 2}y^2.

    And so on....

    Continuing in this way, we get that
    \displaystyle (x + y)^n = \binom{n}{n}x^ny^0 + \binom{n}{n - 1}x^{n - 1}y^1 + ... + \binom{n}{0}x^0y^n = \binom{n}{0}x^0y^n + ... + \binom{n}{n}x^ny^0 =  \sum_{k=0}^{n} \binom{n}{k} x^ky^{n-k}.
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    This set of videos explains it somewhat I believe...
    https://www.khanacademy.org/math/alg...nomial-theorem
    I suggest watching this in conjunction (not at the same time though obvsly.. lol) with reading this...
    http://www.mathsisfun.com/combinator...mutations.html
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    (Original post by 0x2a)
    First of all (x + y)^n = \underbrace{(x + y)(x + y)...(x + y)}_{\text{n times}}.

    Now what you want to do to get the general formula for the binomial theorem (when n \in \mathbb{N}), is to collect all the individual x^r, \ r = 0, 1, ... n together.

    For the x^n, we see that we want n of the xs out of n, so we have \displaystyle \binom{n}{n} choices, which is equal to 1. But we have no ys in our expression, as all of our choices have been filled with xs.

    For the x^{n - 1}, we see that we need to choose n-1 xs out of n ones. Now clearly this is \displaystyle \binom{n}{n - 1}. But if you choose n - 1 xs, you always have 1 y with your x^{n - 1}, so we have that \displaystyle \binom{n}{n - 1}x^{n - 1}y.



    Same logic for x^{n - 2}. We need to choose n - 2 xs out of n, so we have \displaystyle \binom{n}{n - 2} choices. But if you wants the parts of the expression with only x^{n - 2}, then you're going to get y^2 with it, as the remaining 2 choices must be given to y. So we have that \displaystyle \binom{n}{n - 2}x^{n - 2}y^2.

    And so on....

    Continuing in this way, we get that
    \displaystyle (x + y)^n = \binom{n}{n}x^ny^0 + \binom{n}{n - 1}x^{n - 1}y^1 + ... + \binom{n}{0}x^0y^n = \binom{n}{0}x^0y^n + ... + \binom{n}{n}x^ny^0 =  \sum_{k=0}^{n} \binom{n}{k} x^ky^{n-k}.
    +5 for patience alone ...
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    (Original post by TeeEm)
    +5 for patience alone ...
    Thank you, but that was nothing after a whole chapter on determinants
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    (Original post by 0x2a)
    Thank you, but that was nothing after a whole chapter on determinants
    still admirable.
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    Sorry to intrude but I was also wondering on how would you prove the binomial theorem when n is not an integer but rational. If anyone has a proof of this would they mind sharing?
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    (Original post by EmptyMathsBox)
    Sorry to intrude but I was also wondering on how would you prove the binomial theorem when n is not an integer but rational. If anyone has a proof of this would they mind sharing?
    https://www.proofwiki.org/wiki/Binom...nomial_Theorem
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    (Original post by EmptyMathsBox)
    Sorry to intrude but I was also wondering on how would you prove the binomial theorem when n is not an integer but rational. If anyone has a proof of this would they mind sharing?
    When n is not an integer you need to use some fairly heavy machinery from Analysis to prove the result. It's not terribly difficult, but the proof relies on setting up various results about power series, convergence and differentiability, so you're basically asking someone to copy out a chunk of textbook work.

    I would see if you can track down a decent Analysis textbook, or try googling for "proof of binomial theorem" to see if there are links to any decent website proofs
 
 
 
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