- #1

- 642

- 196

[tex]

C^2_2 = 1\:\:\:C^2_3 = 3\:\:\:C^2_4 = 6\:\:\:C^2_5 = 10 \\

We\:can\:see\:that\:\:\:C^2_n = C^2_{n-1} + (n-1),\:let's\:try\:n = 5 \\

C^2_5 = C^2_4 + (5 - 1) \\

= C^2_3 + (4 - 1) + (5 - 1) \\

= C^2_2 + (3 - 1) + (4 - 1) + (5 - 1) \\

= 1 + (3 - 1) + (4 - 1) + (5 - 1) \\

C^2_5 = (1 - 1) + (2 - 1) + (3 - 1) + (4 - 1) + (5 - 1) = 10 \\

Generally,\: C^2_n = (1 - 1) + (2 - 1)\:+\:...\:+\:(n - 1) \\

= (1 + 2\: +\: ... \: + \: n) - n \\

= \frac{(n+1)n}{2} - n \\

= \frac{n^2-n}{2}

[/tex]