For how many integers n where \(2 \le n \le 100\) is \(\binom{n}{2}\) odd?

RektTheNoob
Sep 9, 2017

#1**0 **

a_{2}=2!/(2!0!)=2/2=1

a_{3}=3!/(2!1!)=6/2=3

a_{4}=4!/(2!2!)=24/4=6

a_{5}=5!/(2!3!)=120/12=10

a_{6}=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 a_{n}=2n

100 is the last term, so a_{n}=100

100=2n

n=50

ANSWER: 50 integers

Guest Sep 9, 2017