Your HSC mark is the average of your assessment mark and your exam mark.
Your exam mark is how well YOU do in the exam.
Because you are ranked 1st, your assessment mark will be equal to the highest exam mark in your cohort.
In this question, when it says 'let M be the maximum value of |a_n-1|, |a_n-2| ...'
EDIT: is it saying M is equal in magnitude to the highest coefficient?
Re: HSC 2015 4U Marathon
I would have thought that unless they asked you to use calculus or it had 'hence' in the question (and a previous part involved using calculus) that you would get full marks...
Hi,
I was wondering which HSC years have had the more challenging exams, as I'd like to have a go at them at some stage. I only really know that 2003 was supposed to be very difficult. (If you'd like to say which years have had easier and moderate papers, that would be great as well.)
For this proof, can I do what I did from lines 3 to 4?
\sum_{n=0}^{\infty }\frac{x^{2^{n}}}{1-x^{2^{n+1}}}\\=\lim_{N\rightarrow \infty }\frac{1-x^{2^{N+1}}-1+x}{(1-x)(1-x^{2^{N+1}})}\\=\lim_{N\rightarrow \infty }\frac{x(1-x^{2^{N+1}-1})}{(1-x)(1-x^{2^{N+1}})}\\=\frac{x(1-x^\infty...