Re: HSC 2015 3U Marathon
Can someone do this I still can't do it:
^n $ for all integers $ n \geq 3)
^k )
^k} (k+2)^{k+1} > (k+2)^{k+1} )
 < (k+1)^2 \Rightarrow k^{k+1} (k+1)^{k+1} < (k+1)^{2(k+1)} \\ \Rightarrow \frac{k^{k+1} (k+2)^{k+1}}{(k+1)^k} < (k+1)^{k+2} )
^{k+1} < \frac{k^{k+1} (k+2)^{k+1}}{(k+1)^k} < (k+1)^{k+2} )
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What is important to grasp here is what was my motivation for starting from k(k+2) < (k+1)^2
It wasn't something coming from thin air, all I did was ask:
"ok, I need to get from the assumption, to the inductive step inequality, how can I transform the assumption inequality to get the lower bound of the assumption"
well, I got this:
^k} (k+2)^{k+1} > (k+2)^{k+1} )
Then by doing this, I ask myself the question
"Now, I need to ask, is this new upper bound, lower than the upper bound of the inductive step, if so, then the problem is complete" (*)
That is, I then need to ask whether:
^k} (k+2)^{k+1} < (k+1)^{k+2} )
is true, and it turns out it is with a bit of simplification of powers.
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(Note, this 'method' is simply a heuristic for problem solving, for SOME problems, asking the question (*) is not always helpful)
 Determine the value of $J$. \\ \indent (ii) By dividing the interval from $x=0$ to $x=\pi$ into $n$ equal sub-intervals and using the mid-ordinate approximation, show that an approximate value of $J$ is given by:\\$K_n=2q_n \left [\sin{(q_n)}+\sin{(3q_n)}+\sin{(5q_n)}+...+\sin{((2n-1)q_n)} \right ]$ where $q_n=\frac{\pi}{2n}$. \\ \indent (iii) Starting with the formula of part (ii), and using the result $\sin{\alpha}\,+\,\sin{3\alpha}\,+\,\sin{5\alpha}+...+\sin{\left [\left (2n-1 \right )\alpha \right ]}=\frac{1-\cos{\left (2n\alpha \right )}}{2\sin{\alpha}}$, deduce the exact value of $J$. It may be assumed that $\lim_{h\to0}\frac{h}{\sin{h}}=1$.)
