[tex]1\circ 2 = \frac{1}{2}(1 - 3)(2 - 3) + 3 = \frac{1}{2}\cdot (-2)\cdot (-1) + 3 = \frac{2}{2}+3 = 1 + 3 = 4\\1\circ 2\circ 3 = 4 \circ 3 = \frac{1}{2}(4-3)(3-3) + 3 = 0 + 3 = 3\\1\circ 2\circ 3\circ 4 = 3\circ 4 = 3\\1\circ \cdots \circ 5 = 1\circ \cdots \circ 4\circ 5 = 3 \circ 5 = 3[/tex].
Putem demonstra cu inductie ca [tex]1\circ 2\circ ... \circ 2015 = 3[/tex]
[tex]p(n): 1\circ 2\circ 3\circ ... \circ n = 3, n \geq 3\\p(3): 1 \circ 2\circ 3 = 3(aratat\:mai\:sus)\\p(k)\to p(k+1), k \geq 3\\p(k): 1\circ 2\circ 3\circ \cdots \circ k = 3\\p(k+1): 1\circ 2\circ \cdots \circ k \circ (k+1) = 3\\3 \circ (k+1) = 3\\\frac{1}{2}(3-3)(k+1-3) + 3 = 3\\ 0 + 3 = 3\\ 3 = 3\\Adevarat[/tex]
Astfel [tex]1\circ 2\circ \cdots \circ 2015 = 3[/tex]
