Pascal's Triangle related question

[chimbooze]

Let n and k be positive integers. After calculating several examples, guess a closed formula for:

(n \ 0) + (n + 1 \ 1) + ... + (n + k \ k)

If it helps, this is the formula for the sum of the nth row of the pascal triangle:

(n \ 0) + (n \ 1) + ... (n \ k) = 2^n

(n \ 0) means n choose 0. I couldn't write that in the forum so I had to improvise. Hopefully you know what it means. The "n" is on top and 0 is on the bottom.

 

[Elucidus]

I know that answer, but I am foggy on how to get it.

I will mention that

\sum_{i=0}^{k} \left( _i^{n+i} \right) = \sum_{i=0}^{k} \left(_n^{n+i} \right)

by symmetry. I'm not sure that helps but it's a different angle to investigate.

--Elucidus

P.S. Look into d-simplex numbers too.