Vtgyjfy

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.

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}

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 $ngeq k$ 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 $binom n{k+1}+binom n{krlap{phantom+}}=binom{n+1}{k+1}$, 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 $kleq ileq n$, 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
$$sum_{i=k}^nbinom ik=binom{n+1}{k+1}.
$$

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.

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.