We're just starting Taylor series now, and I got kind of stumped on a problem. Is there a way to find the nth term part of the series, or is it just intuition by looking at some of the terms?
$$f(x)=(1+x)^{-3}.$$
$\sum_{n=0}^\infty\frac{(-1)^n x^n(1 + n)(2 + n)}2$ is the answer but, looking at the terms, the $\frac{1}{2}(n+1)(n+2)$ part doesn't jump out at me. Perhaps I just lack the number sense to do this, but help would be appreciated.
On
We already know that $$(1+x)^{-1}=\frac{1}{1+x}=1-x+x^2-x^3+x^4-x^5+x^6-x^7+\dots,\qquad |x|<1.$$
Applying term-by-term differentiation, we get
$$(-1)(1+x)^{-2}=0-1+2x-3x^2+4x^3-5x^4+6x^5-7x^6+\dots,\qquad |x|<1.$$ So get,
$$(-1)(1+x)^{-2}=-1+2x-3x^2+4x^3-5x^4+6x^5-7x^6+\dots,\qquad |x|<1.$$
Applying again the term-by-term differentiation, we get
$$(-1)(-2)(1+x)^{-3}=0+2-6x+12x^2-20x^3+30x^4-42x^5+\dots,\qquad |x|<1.$$ Simplifying, we get $$\frac{2}{(1+x)^3}=2-6x+12x^2-20x^3+30x^4-42x^5+\dots,\qquad |x|<1.$$ Finally, by multiplying $\frac{1}{2}$ to both sides, we get $$\frac{1}{(1+x)^3}=1-3x+6x^2-10x^3+15x^4-21x^5+\dots,\qquad |x|<1.$$
Observe that $1,3,6,10,15,21,\dots$ are called triangular numbers. Thus,
$$\frac{1}{(1+x)^3}=\sum_{n=1}^\infty(-1)^{n-1}\frac{n(n+1)}{2}x^{n-1},\quad |x|<1$$ Equivalently, we get $$\frac{1}{(1+x)^3}=\sum_{n=0}^\infty(-1)^{n}\frac{(n+1)(n+2)}{2}x^n,\quad |x|<1$$
NOTE: In case you don't know the properties of triangular numbers: Observe the following. $$\begin{align} 1&=1\\ 3&=1+2\\ 6&=1+2+3\\ 10&=1+2+3+4\\ 15&=1+2+3+4+5\\ 21&=1+2+3+4+5+6\\ \end{align} $$ In general, $$\frac{n(n+1)}{2}=1+2+3+\dots+n.$$
For problems of the form $(a+x)^n$, you can actually cheat and use the generalized binomial expansion (which is secretly just Taylor expansions for a binomial to a power)
$$(a+x)^n=\sum_{k=0}^\infty\binom nk a^{n-k}x^k$$
where
$$\binom nk=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$
Which should help make the answer jump out at you as more clear.