Question

CAN WE WRITE (2n-2)! AS 2(n-1)!

Answer 1

Answer:

Note that:

(2n+2)!=(2n+2)⋅(2n+1)⋅2n⋅(2n−1)⋅(2n−2)⋯⋅2⋅1=(2n)!

(2n+2)!=(2n+2)⋅(2n+1)⋅2n⋅(2n−1)⋅(2n−2)⋯⋅2⋅1⏟=(2n)!

Which means

(2n+2)!=(2n+2)⋅(2n+1)⋅(2n)!

(2n+2)!=(2n+2)⋅(2n+1)⋅(2n)!

So when dividing (2n+2)!(2n+2)! by (2n)!(2n)! only those first two factors of (2n+2)!(2n+2)! remain (in this case in the denominator

Hint: One very useful property of factorials is that

Step-by-step explanation:

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Answer 2

Answer:

$\huge\mathfrak\red{hola mates}$

given -

{2n(2n-2)(2n-4)….(2)}*{(2n-1)(2n-3)…..1}

Thus we have n even and n odd terms.

Take 2 common from all the even terms.

2n∗n(n−1)(n−2)....1∗(2n−1)(2n−3)…..12n∗n(n−1)(n−2)....1∗(2n−1)(2n−3)…..1

=2n∗n!∗2n∗n!∗ product of all odd terms upto 2n.

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