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?


$\sum_{n=0}^\infty\frac{(-1)^n x^n(1 + n)(2 + n)}2$ 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.

  • $\begingroup$ Start from the expansion of $$\frac1{1+x}$$ which you should know, and differentiate it twice to get the expansion of $2f(x)$. $\endgroup$ – Did Feb 27 '17 at 0:46

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)

$$(a+x)^n=\sum_{k=0}^\infty\binom nk a^{n-k}x^k$$


$$\binom nk=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$

Which should help make the answer jump out at you as more clear.

  • $\begingroup$ I'm still not getting it. I've written out the first few terms, and, after simplifying it: ${1-3x+6x^2-10x^3+15x^4+...}$ I'm not sure where to go, or how that theorem fits into it $\endgroup$ – addadisplayname Feb 27 '17 at 0:58
  • $\begingroup$ You should note that you get something of the following form: $$\binom{-3}k=\frac{(-3)(-4)(-5)\cdots(-2-k)}{k!}=(-1)^k\frac{3\cdot4\cdot5\cdots(k+2)}{1\cdot2\cdot3\cdots k}=?$$ $\endgroup$ – Simply Beautiful Art Feb 27 '17 at 0:59
  • $\begingroup$ Since my $a=1$, $a^{-3-k}=1$, so it's just the sum of $(-1)^k\frac{k+2}{k}x^k$? $\endgroup$ – addadisplayname Feb 27 '17 at 1:25
  • $\begingroup$ @addadisplayname Actually, you should check how the numbers cancel. Everything in the denominator cancels out except $1\cdot2$. Everything in the numerator cancels except $(k+1)(k+2)$. Thus, we are left with$$\binom{-3}k=\frac{(-1)^k(k+1)(k+2)}2$$And$$(1+x)^{-3}=\sum_{k=0}^\infty\frac{(-1)^k(k+1)(k+2)}2x^k$$ $\endgroup$ – Simply Beautiful Art Feb 27 '17 at 1:38
  • $\begingroup$ Thank you. I wasn't thinking of it in the sense of the actual terms, I just thought of the whole thing in terms of k, now I see where the 1/2 comes from. $\endgroup$ – addadisplayname Feb 27 '17 at 2:02

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, $$\frac{n(n+1)}{2}=1+2+3+\dots+n.$$

  • $\begingroup$ (+1) for the standard way. Personally, I prefer to do it all in sigma notation, which avoids the mention of the triangle numbers. $\endgroup$ – Simply Beautiful Art Feb 27 '17 at 2:15
  • $\begingroup$ Yes, thank you for the tip. --) (+1) to you as well. $\endgroup$ – Juniven Feb 27 '17 at 2:17

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