2018 test 1 variant C 2b

[Maria-Clara Eberlein]

How did we get from the second last to last line, in particular the calculation with (2n)! and (2n+2)! (See attachment)

[RunboZhang]

It will be clearer if you expand the factorial:
 $2n! = 2n \cdot (2n-1) \cdot ... \cdot 1$,
$(2n+2)! = (2n+2) \cdot (2n+1) \cdot (2n) \cdot ... \cdot 1 = (2n+2) \cdot (2n+1) \cdot (2n)!$
Thus $\frac {(2n)!}{(2n+2)!} = \frac {1}{(2n+2) \cdot (2n+1)}$

[bnogas]

Don't we need to have 1/2n(2n+1) not 1/(2n+2)(2n+1) ?

[RunboZhang]

I believe the equation can be written as:
$ = |z| \cdot \frac{(n+1)^{3}}{(2n+2)(2n+1)} $
$ = |z| \cdot \frac{(n+1)^{2}}{2 \cdot (2n+1)}$

[bnogas]

Shouldn't it be 2n(n+1) in the denominator, not 2(n+1)?

[RunboZhang]

No. I think it should be $(2n+2) \cdot (2n+1) $in the denominator. The last line of my equation in previous comment is the same stuff. Since (2n+2)=2(n+1), then cancel out the same term on denominator and numerator. If I didn't make myself clear pls comment below

[bnogas]

Thx, but I meant in the very first post with a screenshot of the official solution it has 1/2n(2n+1),so I was just confused as to which one was the correct answer.

[RunboZhang]

Sorry I didn't not make myself clear on this, I think the answer has a typo. The denominator in the original screenshot should be $(2n+2)\cdot(2n+1) $

[Maria-Clara Eberlein]

Ok if it has a typo then that makes sense, thank you!