Triangle of numbers sum What is the total of all these numbers? 
                  1 
                 1 2
                1 2 3
               1 2 3 4
              1 2 3 4 5
             . . . . . .
            1 2 ........N  

The answer should be a single binomial coefficient. It looks pretty easy, but I cannot come to a solution.

The total of the $1$'s is $N$. The total of the $2$'s is $2(N-1)$, the toal of the $3$'s is $3(N-2)$,..., the total of the $N$'s is $N(N-(N-1))=N$.
Throwing this into a sum we find:
$$\sum_{i=1}^N i \cdot (N-(i-1)) = \sum_{i=1}^N i \cdot N - \sum_{i=1}^N i^2 + \sum_{i=1}^N i$$
$$= N \frac{N(N+1)}{2} - \frac{N(N+1)(2N+1)}{6} + \frac{N(N+1)}{2}$$
$$= \frac{N(N+1)^2}{2} - \frac{N(N+1)(2N+1)}{6}.$$
We can factor this into $$N(N+1) \left( \frac{(N+1)}{2} - \frac{(2N+1)}{6} \right)$$
$$=N(N+1) \left( \frac{N+2}{6} \right) = \frac{N(N+1)(N+2)}{6} = \frac{(N+2)!}{(N-1)!3!} = { N+2 \choose 3 }$$
