How is this a property of Pascal's triangle? For all non-negative integers $k$ and $n$,
$$
\dbinom{k}{k} + \dbinom{k+1}{k} + \dbinom{k+2}{k} + \ldots + \dbinom{n}{k} = \dbinom{n +1}{k+1}
$$
How is this a property of Pascal's triangle? I do not see at all how it relates. I'm looking for an explanation, as opposed to some calculation with algebraic manipulations.
 A: By induction on $n$. If $n<k$ then the left hand side is an empty sum, and the right hand side has $0<n+1<k+1$ so $\binom{n+1}{k+1}=0$. For $n\geq k$ assume the result for $n-1$, so one has
$$
  \dbinom{k}{k} + \dbinom{k+1}{k} + \dbinom{k+2}{k} + \ldots + \dbinom{n-1}{k} =  \dbinom n{k+1}.
$$
Substituting that into the equation to be proved, we need to show $\binom n{k+1}+\binom n{k\rlap{\phantom+}}=\binom{n+1}{k+1}$, which is a familiar identity.
There is also a direct combinatorial interpretation of the formula. If you choose a subset of $k+1$ from the $n+1$ numbers $\{0,1,\ldots,n\}$, the last selected number $i$ satisfies $k\leq i\leq n$, and the remaining $k$ numbers can be chosen in $\tbinom ik$ ways among $\{0,\ldots,i-1\}$. Every subset is accounted for exactly once, so
$$\sum_{i=k}^n\binom ik=\binom{n+1}{k+1}.
$$
A: The "usual" property of the Pascal triangle is $$\dbinom{n+1}{k+1} = \underbrace{\dbinom{n}{k+1}}_{\text{Up right term}} + \underbrace{\dbinom{n}k}_{\text{Up left term}}$$
Now proceed along the up right branch i.e. $$\dbinom{n}{k+1} = \underbrace{\dbinom{n-1}{k+1}}_{\text{Up right term}} + \underbrace{\dbinom{n-1}{k}}_{\text{Up right term}}$$
Now again take the up right branch and keep proceeding to finally end with
$$\dbinom{k+1}{k+1} = \dbinom{k}{k}$$
Now put these together to get what you want.
EDIT
$\hskip2.5in$
\begin{align}
\color{red}{\dbinom{n+1}{k+1}} & = \color{blue}{\dbinom{n}{k}} + \color{red}{\dbinom{n}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{red}{\dbinom{n-1}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \color{red}{\dbinom{n-2}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{red}{\dbinom{k+2}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{blue}{\dbinom{k+1}{k}} + \color{red}{\dbinom{k+1}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{blue}{\dbinom{k+1}{k}} + \color{blue}{\dbinom{k}{k}}
\end{align}
A: To add to Marvis answer note that the first equality in his answer is correct by this combinatorical argument:
When chossing $k+1$ elements out of $n+1$ elements we have two choises (that do not intersect): Either the last (say we number them from $1$ to $n+1) $element is chosen or not.
In the case that the last element is not chosen we have to choose all $k+1$ elements from $n$ elements (since the last one isn't picked) and in the second case we only need to choose another $k$ elements.
A: You're asking, for example, for the sum of all the indicated cells of Pascal's triangle
$$ \begin{matrix} \cdot
\\ \cdot & \cdot
\\ \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & 
\\ \cdot & \cdot & \cdot & \bullet & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot
\end{matrix} $$
which is the same thing as the sum of the cells
$$ \begin{matrix} \cdot
\\ \cdot & \cdot
\\ \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \circ
\\ \cdot & \cdot & \cdot & \bullet & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot
\end{matrix} $$
because I've added in zero. Now, what do you know about the sum of adjacent cells of Pascal's triangle?
Of course, this method is still a calculation with algebraic manipulations, just organized in picture form.
