There's a vast amount of clutter on the internet about this which I've been trawling through but it does not answer exactly what I'm asking which, because superficially similar questions have been on MSE before, I'll try to be very clear about.

Firstly, I've proved The Binomial Theorem for positive integers using only algebra in the standard way as shown here : https://www.math.hmc.edu/calculus/tutorials/binomial_thm/induction.html

I wanted a similarly mathematically unsophisticated level of proof to extend The Binomial Theorem to negative integers. That is without using, for example, Taylor's theorem or devices such as the gamma function.

I'm well aware that the standard approach is to develop Taylor's or McLaurin's Theorem from which the negative integers, rationals and, indeed, real cases follow.

I've come up with what I hope is a minimalist level proof using geometric progressions (to deal with the convergence issue) and simple differentiation up to the level of the chain rule.

As it's still working with integers, it's a proof by induction, but 'downwards' from $n=-1$ to $-\infty$. I think this has not been explored before, as far as I can tell, because proof writers want to tackle rational powers at the same time as negative integer powers, and Taylor's theorem is beckoning.

I'd like to know,

a) Is it a valid proof ? If not, can it be saved ? In particular, is there a simple way to justify the validity of differentiating the infinite series in the inductive step ?

b) Is there any possible simpler proof that uses lower level mathematics ?

I've not seen this anywhere, hence my cautious approach.

Statement: (The Binomial Theorem for negative integers) $$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+...+\frac{n(n-1)(n-2)\dots(n-r+1)}{r!}x^r+...$$ For $n=-1$ the theorem becomes, $$(1+x)^{-1}=1-x+x^2-x^3+x^4-x^5+...$$ The RHS is in geometric progression, with first term, $a=1$ and common ratio, $r=-x$.

Provided $-1<r<1$ the series is convergent and has a sum to infinity of, $$\frac{a}{1-r}=\frac{1}{1+x}$$ This is the LHS and so the theorem is true for $n=-1$ with $-1<x<1$.

Now to establish the inductive step.

Assume the theorem is true for some $n=-k$, $k>1$, $-1<x<1$ in which case, $$(1+x)^{-k}$$ $$=1-kx+\frac{(-k)(-k-1)}{2!}x^2+\dots+\frac{(-k)(-k-1)(-k-2)\dots(-k-r+1)}{r!}x^r+\dots$$ Further assume that it is valid to differentiate both sides with respect to $x$, $$(-k)(1+x)^{-k-1}$$ $$=(-k)+\frac{2(-k)(-k-1)}{2!}x+\dots+\frac{r(-k)(-k-1)(-k-2)\dots(-k-r+1)}{r!}x^{r-1}+\dots$$ Divide through by $(-k)$, $$(1+x)^{-k-1}$$ $$=1+\frac{(-k-1)}{1!}x+\dots+\frac{(-k-1)(-k-2)\dots(-k-r+1)}{(r-1)!}x^{r-1}+\dots$$ Tidy up, $$(1+x)^{-(k+1)}$$ $$=1-(k+1)x+\dots+\frac{(-(k+1))(-(k+1)-1))\dots(-(k+1)-r)}{(r-1)!}x^{r-1}+\dots$$ $$=1-(k+1)x+\dots+\frac{(-(k+1))(-(k+1)-1))\dots(-(k+1)-r+1)}{r!}x^{r}+\dots$$ which is exactly the formula assumed to be true with $k+1$ in place of $k$. (That is, the integer one more negative that the integer before)

Thus the inductive step is proved and The Binomial Theorem is valid for all negative integers, provided $-1\lt x\lt1$

  • 1
    $\begingroup$ I don't offhand see anything wrong with your proof. Regarding the validity of differentiating both sides, I don't know of any simple way to justify it, but I recall it's always valid if the series is absolutely convergent. Regarding your second question, if you consider using combinatorics to be "lower level mathematics", then the proofs at Combinatorial proof of Negative Binomial Identity may be of interest, but you may not necessarily consider them to be "simpler" than your proof. $\endgroup$ – John Omielan Apr 15 at 23:26

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