Why the binomial expansion of $(1+x)^n$, $n$ belongs to negative integer or fraction, is $$ 1+nx+n(n-1)x^2/2........ $$ I admit the expansion when $n$ is positive integer, is because of Pascal triangle.
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Welcome to Math SE. Note Newton's generalized binomial theorem gives a basic explanation of this. – John Omielan Jan 26 '20 at 06:29
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But Pascal triangle is for +ve integers only. – John Horan Jan 26 '20 at 06:36
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The generalised binomial sum is $(1+x)^r = \sum_{k=0}^{\infty}\begin{bmatrix} r \ k \end{bmatrix} x^k$ for $r$ a fraction or negative power. You can prove it by finding the Taylor series of $(1+x)^r$. – fGDu94 Jan 26 '20 at 06:52
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Does this answer your question? Generalized binomial theorem – Simply Beautiful Art Jan 26 '20 at 14:42
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Well, for a negative integers $n$ you just need a simple substitution , for example assume we want to expand $1+x$ for $n=-3$, based on the formula we have:
setting $-3=n$ follows: $$\sum_{k=0}^{∞}{{n}\choose{k}}x^{k}$$$$={{n}\choose{0}}x^{0}+{{n}\choose{1}}x^{1}+{{n}\choose{2}}x^{2}+...$$$$=\frac{n!}{0!n!}x^{0}+\frac{n!}{1!\left(n-1\right)!}x^{1}+\frac{n!}{2!\left(n-2\right)!}x^{2}$$$$=1+nx+\frac{n\left(n-1\right)}{2}x^{2}$$
Wow substitute back:
$$=1-3x+6x^{2}-...$$
Also duo to the convergence we need the upper limit of the sum to be $∞$