Show Pascal triangle properties I need to prove two pascal triangle properties:
1)  $\sum_{k=0}^{n}\binom{p+k}{k}=\binom{p+n+1}{n}$
2) $\sum_{k=0}^{n}\binom{k}{p}=\binom{n+1}{p+1}$
I need some advice on how to approach to this kind of summatorial problems involving binomial coefficients.
 A: Hint: If $$\sum_{k=0}^{n}\binom{p+k}{k}=\binom{p+n+1}{n}$$
then
$$\sum_{k=0}^{n+1}\binom{p+k}{k}=\sum_{k=0}^{n}\binom{p+k}{k}+\binom{p+n+1}{n+1}=$$
$$=\binom{p+n+1}{n}+\binom{p+n+1}{n+1}=\binom{p+(n+1)+1}{(n+1)}$$
We use Pascal identity
$$\binom{m}{k}+\binom{m}{k-1}=\binom{m+1}{k}$$
Similarly for second identity
If$$\sum_{k=0}^{n}\binom{k}{p}=\binom{n+1}{p+1}$$
then$$\sum_{k=0}^{n+1}\binom{k}{p}=\sum_{k=0}^{n}\binom{k}{p}+\binom{n+1}{p}=$$
$$=\binom{n+1}{p+1}+\binom{n+1}{p}=\binom{(n+1)+1}{p+1}$$
A: Here is an approach using negative binomial coefficients:
$$
\begin{align}
\binom{-n}{k}
&=\frac{(-n)(-n-1)(-n-2)\cdots(-n-k+1)}{k!}\\
&=(-1)^k\binom{n+k-1}{k}\tag{1}
\end{align}
$$

Question 1: Applying $(1)$ twice and using Vandermonde's Identity yields
$$
\begin{align}
\sum_{k=0}^n\binom{p+k}{k}
&=\sum_{k=0}^n\binom{p+k}{k}\binom{n-k}{n-k}\\
&=\sum_{k=0}^n(-1)^k\binom{-p-1}{k}(-1)^{n-k}\binom{-1}{n-k}\\
&=(-1)^n\binom{-p-2}{n}\\
&=\binom{n+p+1}{n}\tag{2}
\end{align}
$$

Question 2: Similarly
$$
\begin{align}
\sum_{k=0}^n\binom{k}{p}
&=\sum_{k=0}^n\binom{k}{p}\binom{n-k}{n-k}\\
&=\sum_{k=0}^n\binom{k}{k-p}\binom{n-k}{n-k}\\
&=\sum_{k=0}^n(-1)^{k-p}\binom{-p-1}{k-p}(-1)^{n-k}\binom{-1}{n-k}\\
&=(-1)^{n-p}\binom{-p-2}{n-p}\\
&=\binom{n+1}{n-p}\\
&=\binom{n+1}{p+1}\tag{3}
\end{align}
$$
A: 2) Hint: Since $\binom{p}{p}=1=\binom{p+1}{p+1}$
$\sum_{k=p}^{n}\binom{k}{p}=\binom{p}{p}+\binom{p+1}{p}+\binom{p+2}{p}+\dots+\binom{n}{p}$
$\Rightarrow\sum_{k=p}^{n}\binom{k}{p}=\underbrace{\binom{p+1}{p+1}+\binom{p+1}{p}}+\binom{p+2}{p}+\dots+\binom{n}{p}$
$\Rightarrow\sum_{k=p}^{n}\binom{k}{p}\,\,\,\,=\,\,\,\,\binom{p+2}{p+1}+\binom{p+2}{p}+\dots+\binom{n}{p}$
A: I like to add a combinatorial proof, to these nice answers.
Note that we can write the first identity $\sum_{k=0}^{n}\binom{p+k}{k}=\binom{p+n+1}{n}$ as
$$\sum_{k=0}^{n}\binom{p+k}{p}=\binom{p+n+1}{p+1}. \tag{*}$$
To prove it consider a $(p+n+1)$-element set $X=\{x_1,x_2,\dots,x_p,x_{p+1},x_{p+2},\dots,x_{p+n+1}\}$. The RHS of $(*)$ is the number of $(p+1)$-element subsets of $X$. What about the LHS? Let us expand the LHS,
$$\sum_{k=0}^{n}\binom{p+k}{p}=\binom{p}{p}+\binom{p+1}{p}+\binom{p+2}{p}+\cdots+\binom{p+n+1}{p}\tag{**}$$
It seems we have $(n+1)$ cases: selecting $p$ elements from a $p$-element set, selecting $p$ elements from a $(p+1)$-element set, $\dots$, and at the end selecting $p$ elements from a $(p+n)$-element set. But how on earth these cases can be related to the RHS, i.e. selecting $(p+1)$ elements from the $(p+n+1)$-element set $X$ ?! Here's how: since we want to select $(p+1)$ elements from $X$, so at least one of these $x_{p+1},x_{p+2},\dots,x_{p+n+1}$ will be selected and we have the following cases,
Case 1: Select $x_{p+1}$ and then $p$ elements from
$$\{x_1,x_2,\dots,x_p\};$$
Case 2: Select $x_{p+2}$ and then $p$ elements from
$$\{x_1,x_2,\dots,x_p,x_{p+1}\};$$
Case 3: Select $x_{p+3}$ and then $p$ elements from
$$\{x_1,x_2,\dots,x_p,x_{p+1},x_{p+2}\};$$
$\vdots$
Case n+1: Select $x_{p+n+1}$ and then $p$ elements from
$$\{x_1,x_2,\dots,x_p,x_{p+1},\dots,x_{p+n}\};$$
Now by addition principle the number of ways to select $(p+1)$ elements from $X$ is
$$\binom{p}{p}+\binom{p+1}{p}+\binom{p+2}{p}+\cdots+\binom{p+n+1}{p}$$
and it's the LHS, so we have $(*)$.

I know it seems long! But, I believe it's worthy and this kind of proof gives you some visions to construct new identities, it lets you to feel the identity, things that you can not find in an algebraic proof.
