• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

HSC 2016 MX2 Marathon ADVANCED (archive) (1 Viewer)

Status
Not open for further replies.

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2016 4U Marathon - Advanced Level

Perhaps something easier (with a bit of guidance), this is possible with 3U knowledge and 4U polynomial knowledge:



 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2016 4U Marathon - Advanced Level

Holy shit I was that close to the actual solution. I got the integral part but from there everything went haywire.
In the future feel free to post your progress, in this case it was just a computation and the integral is not essential to understanding the problem
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2016 4U Marathon - Advanced Level

Perhaps an interesting excercise:

 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: HSC 2016 4U Marathon - Advanced Level



That's as far as I could get. I know that it must be shown to be always true given the conditions y >= x > 1, but I'm not sure how.


 
Last edited:

lita1000

Member
Joined
Aug 29, 2015
Messages
64
Gender
Female
HSC
2015
Re: HSC 2016 4U Marathon - Advanced Level

does any 2016ers need solutions for sy's past 3 questions before his most recent? I can quickly type them up if requested!
 

lita1000

Member
Joined
Aug 29, 2015
Messages
64
Gender
Female
HSC
2015
Re: HSC 2016 4U Marathon - Advanced Level

oh nvm, just realised sy already posted solutions for the polygon and average distance question, no need to post those ones up then. If a 2016ers needs soln for the altitude q, I can type them up.
 

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,556
Location
Outside reality
Gender
Male
HSC
2016
Re: HSC 2016 4U Marathon - Advanced Level

oh nvm, just realised sy already posted solutions for the polygon and average distance question, no need to post those ones up then. If a 2016ers needs soln for the altitude q, I can type them up.
yes please.
 

dan964

what
Joined
Jun 3, 2014
Messages
3,479
Location
South of here
Gender
Male
HSC
2014
Uni Grad
2019
Re: HSC 2016 4U Marathon - Advanced Level

First...
Case 0: trivial, Case 1: LHS = 6, which is divisible by 6,
Case 2: LHS = 30, which is divisible by 6
Assume formula is true up to n=k:
k(k+1)(2k+1)=6M, M is a natural number
For n=k+1
LHS = (k+1)(k+2)(2k+3)
= (k+1)(2k^2 + 7k + 6)
= k(k+1)(2k + 1+ 6) + 6(k+1)
= k(k+1)(2k + 1) + 6k(k+1)+6(k+1)
= 6M + 6(k+1)^2
= 6K, K is a natural number
Since true for first case, and also for two consecutive cases, it is therefore true for all natural n


am I right to assume that it is asking for 12 is a factor of the n^4-n^2
Case n=0 or n=1
n^4-n^2 = 0, trivial
Case n=2
n^4-n^2 = 4*3 = 12, which is divisible by 12

Assume formula to be true, up to some natural number 'k'
k^4-k^2=12N where N is a natural number (not 0 or 1)

For n=k+1
LHS = (k+1)^4 - (k+1)^2
= k^4 + 4k^3 + 6k^2 + 4k+1 - k^2 - 2k -1
= 12N + 4k^3 + 6k^2 + 2k
= 12N + 2k(2k^2 + 3k+1)
= 12N + 2k(2k+1)(k+1)
It remains to show that n(n+1)(2n+1) is a multiple of 6, which was conducted earlier
= 12N + 2k*6Z (Z a natural number)
= 12(N+kZ)
Hence since the first case, two consecutive cases are true, the statement is proved true for all natural n.
 

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,556
Location
Outside reality
Gender
Male
HSC
2016
Re: HSC 2016 4U Marathon - Advanced Level

Which part do you specifically need help with(as in first part or second part)? I'll type up as soon as I come back from work
second part
First...
Case 0: trivial, Case 1: LHS = 6, which is divisible by 6,
Case 2: LHS = 30, which is divisible by 6
Assume formula is true up to n=k:
k(k+1)(2k+1)=6M, M is a natural number
For n=k+1
LHS = (k+1)(k+2)(2k+3)
= (k+1)(2k^2 + 7k + 6)
= k(k+1)(2k + 1+ 6) + 6(k+1)
= k(k+1)(2k + 1) + 6k(k+1)+6(k+1)
= 6M + 6(k+1)^2
= 6K, K is a natural number
Since true for first case, and also for two consecutive cases, it is therefore true for all natural n


am I right to assume that it is asking for 12 is a factor of the n^4-n^2
Case n=0 or n=1
n^4-n^2 = 0, trivial
Case n=2
n^4-n^2 = 4*3 = 12, which is divisible by 12

Assume formula to be true, up to some natural number 'k'
k^4-k^2=12N where N is a natural number (not 0 or 1)

For n=k+1
LHS = (k+1)^4 - (k+1)^2
= k^4 + 4k^3 + 6k^2 + 4k+1 - k^2 - 2k -1
= 12N + 4k^3 + 6k^2 + 2k
= 12N + 2k(2k^2 + 3k+1)
= 12N + 2k(2k+1)(k+1)
It remains to show that n(n+1)(2n+1) is a multiple of 6, which was conducted earlier
= 12N + 2k*6Z (Z a natural number)
= 12(N+kZ)
Hence since the first case, two consecutive cases are true, the statement is proved true for all natural n.
Alternatively, prove that if n is true, n+6 is true, then show trivially that it is true for the first six natural numbers and draw your conclusion from there.
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top