For how many integers n where \(2 \le n \le 100\) is \(\binom{n}{2}\) odd?
a2=2!/(2!0!)=2/2=1
a3=3!/(2!1!)=6/2=3
a4=4!/(2!2!)=24/4=6
a5=5!/(2!3!)=120/12=10
a6=6!/(2!4!)=720/48=15
The pattern goes 1,3,6,10,15,21,28,36,45,55,etc. or odd,odd,even,even,odd,odd,even,even,etc.
1+2=3
3+3=6
6+4=10
10+5=15
15+6=21
...
(n choose 2) is odd when n = 2,3,6,7,10,11,14,15,18,19...94,95,98,99.
We can add one to every other term (starting with 3) to make a more comprehendable pattern: 2,4,6,8...94,96,98,100. The number of terms stays the same.
This sequence follows the rule an=2n
100 is the last term, so an=100
100=2n
n=50
ANSWER: 50 integers