Question: I want to sketch $|\sqrt{x}-5|$. Now I know what it is going to look like using basic graphing tools. 

My question is, is there an intuitive way to figure out what the graph looks like without making a table of values or using technology to see what the graph looks like?

I know $\sqrt{x}-5$ shifts the graph down five units, but I don't understand why the absolute value flips the graph.

Answer: 
For $|x|<1,$ we have $$\sum\_{n=0}^\infty x^n=\frac{1}{1-x}.$$ Now take $k$ derivatives on both sides: $$\sum\_{n=k}^\infty\frac{n!}{(n-k)!}x^{n-k}=\frac{k!}{(1-x)^{k+1}}.$$ Thus, $$\sum\_{n=k}^\infty\frac{n!}{(n-k)!k!}x^{n}=\frac{x^k}{(1-x)^{k+1}}.$$ Since $\binom{n}{k}=\frac{n!}{k!(n-k)!},$ $$\sum\_{n=k}^\infty\binom{n}{k}x^{n}=\frac{x^k}{(1-x)^{k+1}}.$$
You can prove it by induction on the formal power series too. The result is clear for $k=0$, in which case you have the geometric series $\sum\_{n=0}^\infty x^n = \frac{1}{1-x}$. 

For simplicity we will write the sum over the whole integer $n\in \mathbb{Z}$, with the classical convention (see [generatingfunctionology](https://www.math.upenn.edu/~wilf/DownldGF.html)
 by H. Wilf) that $\binom{n}{j}$ is $0$ if $j>n$ or $j < 0$.

If we take the result to be true for $k$, for $k+1$ we have
$$
(1-x)\sum\_n \binom{n}{k+1} x^n = \sum\_n \binom{n}{k+1}x^n - \sum\_n \binom{n}{k+1}x^{n+1}
$$
and the second sum can be rewritten in terms of $n\mapsto n+1$ to get
$$
(1-x)\sum\_n \binom{n}{k+1} x^n = \sum\_n \binom{n}{k+1}x^n - \sum\_n \binom{n-1}{k+1}x^n
$$
which means that the coefficients of $(1-x)\sum\_n \binom{n}{k+1} x^n$ are
$$
[x^j] \left\{(1-x)\sum\_n \binom{n}{k+1} x^n\right\} = \binom{j}{k+1}-\binom{j-1}{k+1}
$$
which equals $\binom{j-1}{k}$ by the classical [Pascal's rule](https://en.wikipedia.org/wiki/Pascal%27s_rule) $\binom{j}{k+1}=\binom{j-1}{k+1}+\binom{j-1}{k}$.

Thus
$$
(1-x)\sum\_n \binom{n}{k+1} x^n = \sum\_n \binom{n-1}{k}x^n = x\sum\_n \binom{n}{k}x^n = x\frac{x^{k}}{(1-x)^{k+1}}\,,
$$
where we have set $n\mapsto n-1$ in the second equality, and used the inductive hypothesis in the third. Now the result follows upon dividing by $1-x$.

**Moral of the story** Intuitively, the identity $$\sum\_n \binom{n}{k}x^n=\frac{x^{k}}{(1-x)^{k+1}}$$ is a *substractive* application of Pascal's rule, while you can try and see that $$\sum\_k \binom{n}{k}x^k = (1+x)^n$$ would be the *additive* one.