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# For how many integers n where is ​ odd?

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

RektTheNoob  Sep 9, 2017
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#1
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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