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 sum of the $k$-th row is $\sum_{h=1}^k h=\frac{k(k+1)}{2}$.
Recalling that $\sum\limits_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$, we have that the sum of the numbers in the triangle is $$\sum_{k=1}^n \left(\frac{k^2}{2}+\frac{k}{2}\right)=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}=\\=\frac{n(n+1)(2n+4)}{12}=\frac{n(n+1)(n+2)}{6}=\\=\binom{n+2}{3}$$
