Question

What's going on in the numerator? Please have a look at the attachment.

Answer 1

Hi

Answer 2

Hi sir

Answer 3

(2n)! = (2n)(2n-1)(2n-2)(2n-3)......5.4.3.2.1

Answer 4

Notice that the alternate terms (2n), (2n-2), ... 4, 2 are even.

Answer 5

Each of them have a factor of 2, yes ?

Answer 6

Got it.. .and then?

Answer 7

simply group even and odd terms first

Answer 8

(2n)! = (2n)(2n-1)(2n-2)(2n-3)......5.4.3.2.1
= [(2n)(2n-2) ... 4.2][(2n-1)(2n-3)...5.3.1]

Answer 9

fine with above step ?

Answer 10

Yes! That's it. It's clear now Thank you

Answer 11

yw :)

Answer 12

But I'm a little confused with the 2^n.

Answer 13

Yeah, let me ask you a question.

Answer 14

6! = 6.5.4.3.2.1

How many even terms are there on the right hand side ?

Answer 15

3

Answer 16

notice that 6 = 2*3

Answer 17

How about 8! ?

Answer 18

How many even terms will be there ?

Answer 19

That would be 4 (8=2*4) am i right?

Answer 20

Yes, see the pattern ?

In `(2n)!` there are `2n` terms, out of which `n` are even and `n` are odd

Answer 21

All Clear.. Thank you very much.

Answer 22

Np :)