MAT334--2020F > Test 1

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)?

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