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Derivation of 1+nx+ n(n-1)x^2/2!... + rep watch

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    Hi I have been told to look at ncr and do some cancelling but would like to see how the binomial expansion of(1+x)^n is derived. i.e derivation of 1+nx+ n(n-1)x^2/2!...
    Thanks!
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    Combinatorial proof of binomial theorem:

    https://www.math.hmc.edu/calculus/tu...inatorial.html

    You could also use Taylor series to prove it easily, but that might be a bit above your level i.e. its 'FP' stuff.
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    (Original post by VannR)
    Combinatorial proof of binomial theorem:
    https://www.math.hmc.edu/calculus/tu...inatorial.html
    You could also use Taylor series to prove it easily, but that might be a bit above your level i.e. its 'FP' stuff.
    You must be the king of unintentionally offensive lol.
    I wasn't really looking for this but I just figured it out for myself .
    ncr= n choose r =n!/[(n-r)!r!]. The formula just is the same as the normal expansion just with the coefficient being n choose r- for example nc2= n(n-1)/2!. I get it now lol. #epicbrainderp
    How would one go about it with Taylor series? (I do understand Fp2 stuff)
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    (Original post by MathMeister)
    You must be the king of unintentionally offensive lol.
    I wasn't really looking for this but I just figured it out for myself .
    ncr= n choose r =n!/[(n-r)!r!]. The formula just is the same as the normal expansion just with the coefficient being n choose r- for example nc2= n(n-1)/2!. I get it now lol. #epicbrainderp
    How would one go about it with Taylor series? (I do understand Fp2 stuff)
    Apologies; I assumed that this proof was taught alongside binomial theorem in C2, and as a result I thought that you were still learning it. I wasn't trying to be rude .

    Let y = (1 + x)^n, and expand using Taylor's expansion (or Maclaurin's expansion - pick your poison). The result will be what you get from the binomial theorem.
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    (Original post by VannR)
    Apologies; I assumed that this proof was taught alongside binomial theorem in C2, and as a result I thought that you were still learning it. I wasn't trying to be rude .

    Let y = (1 + x)^n, and expand using Taylor's expansion (or Maclaurin's expansion - pick your poison). The result will be what you get from the binomial theorem.
    Hehe that's alright.
    I will try this
    Thank you!
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    (Original post by MathMeister)
    Hi I have been told to look at ncr and do some cancelling but would like to see how the binomial expansion of(1+x)^n is derived. i.e derivation of 1+nx+ n(n-1)x^2/2!...
    Thanks!
    If you want a different way, you can do it by induction: either directly, if you're careful about keeping track of the coefficients, or (slightly more weirdly) by differentiating, expanding inductively, and then integrating again.
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    (Original post by Smaug123)
    If you want a different way, you can do it by induction: either directly, if you're careful about keeping track of the coefficients, or (slightly more weirdly) by differentiating, expanding inductively, and then integrating again.
    Wow.
    btw what is expanding inductively?
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    (Original post by MathMeister)
    Wow.
    btw what is expanding inductively?
    If you already have the expansion for (1+x)^n, you can get the expansion for (1+x)^(n+1) by multiplying (that expansion) by (1+x).

    Edit: (well, that's what I would expect it to mean, but reading the post I'm not so sure that's what he *does* mean..)
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    (Original post by MathMeister)
    Wow.
    btw what is expanding inductively?
    (Original post by DFranklin)
    If you already have the expansion for (1+x)^n, you can get the expansion for (1+x)^(n+1) by multiplying (that expansion) by (1+x).

    Edit: (well, that's what I would expect it to mean, but reading the post I'm not so sure that's what he *does* mean..)
    That is what I mean: (1+x)^{n-1} (1+x) = \left( x^{n-1} + (n-1) x^{n-2} + \dfrac{(n-1)(n-2)}{2} x^{n-3} + \dots + (n-1)x + 1 \right)(1+x) = 1+x(1+n-1) + x^2 \left(\dfrac{(n-1)(n-2)}{2} + (n-1) \right) + \dots.

    The differentiation way: \dfrac{d}{dx} \left[ (1+x)^n \right] = n (1+x)^{n-1} = n \left(1+(n-1)x+\dots + (n-1)x^{n-2}+x^{n-1} \right), which we may integrate if we're careful to keep track of our coefficients.
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    (Original post by Smaug123)
    That is what I mean: ...
    Yes, that's what I expected the "expanding inductively" to mean.

    But when you said "expanding inductively" it seemed to be in relation to the differentiation method, which was what I didn't follow. (I think you perhaps put those words in slightly the wrong place).
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    (Original post by DFranklin)
    Yes, that's what I expected the "expanding inductively" to mean.

    But when you said "expanding inductively" it seemed to be in relation to the differentiation method, which was what I didn't follow. (I think you perhaps put those words in slightly the wrong place).
    Ah, sorry.
 
 
 
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