Binomial expansion of negative exponent

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Stuck on this binomial result doing gravitation chapter in physics the expression is $$ (1+x)^{-2}=1-2x $$ provided $x$ is so smaller than $1$ . My questions is

Why and second If I want to study on my own from where should I study this result with some basic text on algebra

Pls provide me reference

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Analogous to the binomial expansion of $(1+x)^n$ where $n$ is a positive integer, which for reference is

$$(1+x)^n = \sum_{k=0}^n \binom n k x^k$$

we can define an expansion in terms of an infinite series for $n$ a negative integer. This is known as the generalized binomial series; a derivation is here. Below, let $n$ be a positive integer, and $|x|<1$. Then,

$$\frac{1}{(1+x)^n} = (1+x)^{-n} = \sum_{k=0}^\infty \binom {s+k-1}{k}(-1)^k x^k$$

Take $x$ small, i.e. $|x| \ll 1$; then, heuristically, $x^k$ for $k>1$ is negligible because it will be much smaller than $x$, ever moreso as $k$ grows even larger. If we take just the first two terms of the expansion, as is fairly common, then

$$\frac{1}{(1+x)^n} \approx \binom{s-1}{0}(-1)^0 x^0 + \binom{s}{1}(-1)^1x^1 = 1 - sx$$

Take $s=2$ for your desired answer. Of course, you can also take more terms if you want a more accurate approximation.

As for where this would pop up ... well, at least in my case, this expansion was noted in my combinatorics class, and proved particularly useful for generating functions. There are probably many other places where this would pop up of course, but I'm mostly speaking from my own experience as a student.

EDIT: As noted in other answers, you could also derive the generalized binomial series by an approach of Taylor series, expressing a function $f$ through its derivatives. This is something that could be accomplished with basic tools typically taught in a first year of calculus.

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That's the first two terms of the Taylor series
$(1 + x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 ...$
which for small positive x < 1 is approximately 1 - 2x.

Taylor series is a topic of calculus.

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Calculate Binomial coefficients as $${\nu \choose k}= \frac{\nu(\nu-1)(\nu-2)....(\nu-k+1)}{k!}.$$ so $${\nu \choose 1}=\nu, {\nu \choose 2}= \frac{\nu(\nu-1)}{2}, {\nu \choose 3}=\frac{\nu(\nu-1)(\nu-2)}{3!}.....$$ Finally in case of $$(1+x)^{-2}={-2 \choose 0} +{-2 \choose 1}x +{-2 \choose 2} x^2+{-2 \choose 3}x^3+....$$ You get $$(1+x)^{-2}=1-2x+3x^2-4x^3+...~~\mbox {ad inf}~~ \mbox{if}~ |x|<1.$$