Re: HSC 2014 4U Marathon - Advanced Level
Here is my attempt
edit: damn just realized there is an error on like 3rd last line, how do i prove l is the positive case?
Re: MX2 Integration Marathon
Yeh the formula should work even for n>=3 but to end up with and integral we can easily evaluate, we would want to sub a number in the form n=3k+2
Re: MX2 Integration Marathon
On the 8th line u made a tiny mistake, 1+2(n-2)/9=(2n+5)/9 instead of (2n+3)/9
As for the n greater than equal to 5, it is not as easy to integrate I_1 as compared to integrating I_2, thats probably why.
Re: HSC 2014 4U Marathon - Advanced Level
Managed to get rid of trial and error to:
Since m and n are integers m+n=4k (1) and (m+n)^2-mn=49k (2) (k>0)
Subbing (1) into (2):
16k^2-m(4k-m)=49k
M^2-4mk+16k^2-49k=0
M=2k+(49k-12k^2)^0.5
N=2k-(49k-12k^2)^0.5
Discriminant must be positive (and...
Re: HSC 2014 4U Marathon
This is what i thought of:
We know that m+n and mn is an integer and hence (m+n)^2-mn (the denominator) is an integer
So the expression can be written in the form p/q where p and q are integers.
Now p/q=4/49=8/98=12/147=16/196... (multiplying top and bottom by...