Understanding why binomial expansions for negative integers produce infinite series I am a mature student self-studying a UK Math A-Level. I have been studying math on and off in my spare time for the past 6 years, so I have some grounding in the subject but it's all a bit piecemeal and disjointed. Hoping doing this A-Level will pull it all together more. I have a decent grasp of most areas up to differential calculus, but there are gaps and some areas are still fuzzy.
I am working through a revision textbook and am currently stuck on binomial expansions for negative integers. I feel like I have a good grasp of binomial expansions for positive integer powers - the binomial theorem for positive integer powers makes intuitive sense to me in terms of combinatorics. But I just can't seem to get an intuitive understanding for why the expansion of a binomial raised to a negative power should be an infinite series.
For example, one example question in my textbook says "find the binomial expansion of $(1+x)^{-2}$, up to and including the term in $x^3$. The $x^3$ term is specified because otherwise the expansion could go on forever".
I must be missing something somewhere. As I understand the exponent rules, $x^{-n} = \frac 1{x^n}$. Therefore I can't see why the answer to the above question wouldn't simply be: $$\frac 1{1+2x+x^2}$$
I have scoured the internet for somewhere to provide an intuitive explanation for why a binomial raised to a negative power should have an infinite expansion, but everything I have come across so far simply states something like "for powers that are not a positive integer, the binomial expansion is an infinite series", without explaining why. Can anyone please help me make sense of this, or point to a resource that provides an intuitive proof/ explanation?
 A: The expansion has the implied meaning of "as a polynomial in $x$". It is true that
$$(1+x)^{-2}=\frac1{1+2x+x^2}$$ but this does not have the desired shape.
Now a function like $(1+x)^{-2}$ has a vertical asymptote at $x=-1$, and this cannot be modelled by a polynomial, which keeps finite values. To obtain this effect, the polynomial must be of infinite degree, so to say (it is in fact an entire series), which gives it the possibility to diverge to infinity.

Without being aware of the generalized binomial formula, you could solve by the method of indeterminate coefficients. Let the entire series
$$a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots$$ and let us establish
$$(1+2x+x^2)\sum_{k=0}^\infty a_kx^k=1.$$
By expanding the product and regrouping the terms of equal degree, the LHS becomes
$$a_0+(2a_0+a_1)x+(a_0+2a_1+a_2)x^2+(a_1+2a_2+a_3)x^3+(a_2+2a_3+a_4)x^4+\cdots$$
and if we set
$$a_0=1,a_1=-2,a_2=3,a_3=-4,a_4=5,\cdots$$ by cancelling all coefficients but the first, we get an "identity" with the RHS. So
$$(1+x)^{-2}=1-2x+3x^2-4x^3+5x^4+\cdots$$ holds when the series converges.

Note that this formula is unable to represent the given function "past" the asymptote (when $|x|\ge1$), and this is the rule with entire series.
In green, the true function, in blue the fourth degree approximation.

