Pascal Triangle general formula

I'm working on a presentation on the Binomial Theorem for my Algebra 2 class and while writing Pascal's Triangle, I came across one of the properties that I haven't seen in a while. That being $$\sum_{k=1, k-j\ge0}^n\binom{k}{k-j}=\binom{n+1}{n-j}$$ I tried playing around with a proof and I'm having trouble because of the arbitrary value for $j$. I also have tried to simplify it so that I don't need the condition under the summation, that being $k-j\ge0$. For example, for $j=1, n=3$ $$1$$$$\color{red}{1} . . .1$$$$1...\color{red}{2}...1$$$$1...3...\color{red}{3}...1$$$$1...4...\color{green}{6}...4...1$$ $$\color{red}{\binom{1}{0}}+\color{red}{\binom{2}{1}}+\color{red}{\binom{3}{2}}=\binom{3+1}{3-1}=\color{green}{\binom{4}{2}}$$ However, if say $j=3$, then the first terms are negative, hence undefined, hence the need for the restriction. So is there are way to get rid of the restriction for a generalized formula?

EDIT: and no, the christmas tree theme was not intentional...but it is festive...

• Consider them as zeros. Again from the Pascal triangle assume that it's surrounded by zeros, the formulas will be still valid. In your example $\binom{4}{0} = \binom{3}{0} + \binom{3}{-1}+\binom{3}{-2}$ – karakfa Dec 9 '13 at 16:00

$$\sum_{k=1, k-j\ge0}^n\binom{k}{k-j}=\binom{n+1}{n-j} \iff \sum_{k=j}^n \binom{k}{j}= \binom{n+1}{j+1}$$
This last identity can be shown using $\displaystyle {n\choose r}={n-1\choose r-1}+{n-1\choose r}$ applied repeatedly:
• very nice. I've seen that identity as well and 1) never took the time to understand it in regards to the triangle 2) should have let the restriction be $k=j$. Thanks for the help! – Eleven-Eleven Dec 9 '13 at 16:19