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

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1. 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!
2. 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.
3. (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)
4. (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.
5. (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!
6. (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.
7. (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?
8. (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..)
9. (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: .

The differentiation way: , which we may integrate if we're careful to keep track of our coefficients.
10. (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).
11. (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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