Toronto Math Forum
Welcome,
Guest
. Please
login
or
register
.
1 Hour
1 Day
1 Week
1 Month
Forever
Login with username, password and session length
News:
Home
Help
Search
Calendar
Login
Register
Toronto Math Forum
»
MAT334--2020F
»
MAT334--Tests and Quizzes
»
Test 1
»
2018 test 1 variant C 2b
« previous
next »
Print
Pages: [
1
]
Author
Topic: 2018 test 1 variant C 2b (Read 527 times)
Maria-Clara Eberlein
Jr. Member
Posts: 11
Karma: 0
2018 test 1 variant C 2b
«
on:
October 12, 2020, 05:16:22 PM »
How did we get from the second last to last line, in particular the calculation with (2n)! and (2n+2)! (See attachment)
Logged
RunboZhang
Sr. Member
Posts: 51
Karma: 0
Re: 2018 test 1 variant C 2b
«
Reply #1 on:
October 13, 2020, 09:45:35 AM »
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 $\frac {(2n)!}{(2n+2)!} = \frac {1}{(2n+2) \cdot (2n+1)}$
Logged
bnogas
Newbie
Posts: 4
Karma: 0
Re: 2018 test 1 variant C 2b
«
Reply #2 on:
October 13, 2020, 11:34:38 AM »
Don't we need to have 1/2n(2n+1) not 1/(2n+2)(2n+1) ?
Logged
RunboZhang
Sr. Member
Posts: 51
Karma: 0
Re: 2018 test 1 variant C 2b
«
Reply #3 on:
October 13, 2020, 12:03:03 PM »
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)}$
Logged
bnogas
Newbie
Posts: 4
Karma: 0
Re: 2018 test 1 variant C 2b
«
Reply #4 on:
October 13, 2020, 12:27:23 PM »
Shouldn't it be 2n(n+1) in the denominator, not 2(n+1)?
Logged
RunboZhang
Sr. Member
Posts: 51
Karma: 0
Re: 2018 test 1 variant C 2b
«
Reply #5 on:
October 13, 2020, 01:00:11 PM »
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
«
Last Edit: October 13, 2020, 01:12:51 PM by RunboZhang
»
Logged
bnogas
Newbie
Posts: 4
Karma: 0
Re: 2018 test 1 variant C 2b
«
Reply #6 on:
October 13, 2020, 01:37:31 PM »
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.
Logged
RunboZhang
Sr. Member
Posts: 51
Karma: 0
Re: 2018 test 1 variant C 2b
«
Reply #7 on:
October 13, 2020, 01:51:14 PM »
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) $
Logged
Maria-Clara Eberlein
Jr. Member
Posts: 11
Karma: 0
Re: 2018 test 1 variant C 2b
«
Reply #8 on:
October 14, 2020, 03:53:24 PM »
Ok if it has a typo then that makes sense, thank you!
Logged
Print
Pages: [
1
]
« previous
next »
Toronto Math Forum
»
MAT334--2020F
»
MAT334--Tests and Quizzes
»
Test 1
»
2018 test 1 variant C 2b