Compute

If a diagonal of numbers of any length is selected starting at any of the 1's bordering the sides of the triangle and ending on any number inside the triangle on that diagonal, the sum of the numbers inside the selection is equal to the number below the end of the selection that is not on the same diagonal itself. If you don't understand that, look at the drawing. 1+6+21+56 = 84 1+7+28+84+210+462+924 = 1716 1+12 = 13

see: http://ptri1.tripod.com/

$$\small{\text{$\left( \begin{array}{c}4 \\ 0 \end{array} \right)$}} +\small{\text{$\left( \begin{array}{c}5 \\ 1 \end{array} \right)$}} +\small{\text{$\left( \begin{array}{c}6 \\ 2 \end{array} \right)$}} +\small{\text{$\left( \begin{array}{c}7 \\ 3 \end{array} \right)$}} =\small{\text{$\left( \begin{array}{c}8 \\ 3 \end{array} \right)$}}\\\\ \small{\text{ $\left( \begin{array}{c}8 \\ 3 \end{array} \right) =\dfrac{8!}{3!\cdot 5!} = \dfrac{\not{5!}\cdot 6 \cdot 7 \cdot 8 }{3!\cdot \not{5!}} = \dfrac{6 \cdot 7 \cdot 8 }{3!} = \dfrac{6 \cdot 7 \cdot 8 }{2\cdot 3} =\dfrac{ 6 } { 3 }\cdot 7 \cdot \dfrac{ 8 } { 2 } = 2\cdot 7 \cdot 4 \\ = 56 $}}$$