+0  
 
+1
248
1
avatar+376 

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

RektTheNoob  Sep 9, 2017
 #1
avatar
0

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

Guest Sep 9, 2017

2 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.