Search results

Re: HSC 2015 3U Marathon This is simply a completion of kawaiipotato's answer, starting from the bold (*) here:

Re: HSC 2015 3U Marathon As kawaiipotato has proven, one needs to only prove: \frac{(k+1)^{k+1}}{2^k} \leq \left(\frac{k+2}{2} \right)^{k+1} We can re-arrange it and cancel out the denominators, to proving: \\ $Prove that$ \ \left(\frac{k+2}{k+1} \right)^{k+1} > 2 \\ $So,$ \...

Re: HSC 2015 3U Marathon It wouldn't receive full marks unless you prove that last line as it isn't an immediate result I'll post my solution soon

Re: HSC 2015 3U Marathon The addition here is the wrong way around

Re: HSC 2015 3U Marathon \\ $i) Using the definition$ \ \binom{n}{r} := \frac{n!}{(n-r)!r!} \\ $So,$ \ \binom{n}{r} = \frac{n!}{(n-r)! r!} = \frac{n!}{(n- (n-r))! (n-r)!} = \binom{n}{n-r} \\ $ii) Transform the expression$ \ f(r) \ $into an easier to deal form, that is, replace every$ \...

Re: HSC 2015 3U Marathon On track up to this part, your proof of this is not valid Just because a \geq b \ $and$ \ c \geq d does not mean \frac{a}{c} \geq \frac{b}{d} (which is what you do when you divide the inequalities (k+2) > 3 and (k+1) > 2) For instance, 4 > 3 and 3 > 2 but...

Re: HSC 2016 4U Marathon - Advanced Level 2014 (actually 2013) stopped because mods wanted it to (I wanted it to keep going)

Re: HSC 2016 4U Marathon - Advanced Level \\ $Show that the only pairs of positive integers$ \ x , y \ $so that$ \\ x^y = y^x \ $and$ \ x \neq y \\ $are$ \ x = 2, y = 4 \ $and$ \ x = 4, y = 2

Post questions within the scope of Mathematics Extension 2 that are in general Q16 and beyond, focusing on problem solving and neat results within the reach of elementary mathematics.

Re: HSC 2015 3U Marathon \\ $Prove by induction or otherwise for integers$ \ n \geq 1 \ $that$ \ n! \leq \left(\frac{n+1}{2} \right)^n

Re: HSC 2015 3U Marathon \\ 3^4 > 4^3 \ $so it is true for$ \ n = 3 \\ $Assume true for$ \ n = k \Rightarrow k^{k+1} > (k+1)^k \\ $Using the inequality from the assumption, we get$ \\\\ \frac{k^{k+1}}{(k+1)^k} (k+2)^{k+1} > (k+2)^{k+1} \\ k(k+2) < (k+1)^2 \Rightarrow k^{k+1} (k+1)^{k+1}...

Anyone with the paper please PM me

Re: HSC 2015 4U Marathon Good luck MX2 students!

Re: HSC 2015 4U Marathon \\ $i)$ \ P_1 = \frac{1}{2} = \frac{c_1}{d_1} = \frac{\text{odd}}{\text{even}} \ \Rightarrow \ $true for$ \ n = 1 \\ $Assume true of$ \ P_k = \frac{c_k}{d_k} \\ P_{k+1} = P_k + \frac{1}{p_{k+1}} = \frac{c_k p_{k+1} + d_k}{d_k p_{k+1}} \\ $Since$ \ p_{k+1} \ $is...

Re: HSC 2015 4U Marathon \\ $Consider the sum$ \ P_n = \sum_{i=1}^n \frac{1}{p_i} \ $where$ \ p_i \ $is the$ \ i^{th} \ $prime number, so$ \\ p_1 = 2, \ p_2 = 3, \ p_3 = 5, \ $etc.$ \\ $i) Prove by mathematical induction that$ \ P_n \ $when simplified into a single fraction, is$ \ P_n =...

Re: HSC 2015 4U Marathon For a HSC solution without invoking AM-GM: \\ $Let$ \ A = x_1 + x_2 + \dots + x_{n-1} \ $and$ \ B = x_1 x_2 \dots x_{n-1} \\ $The inequality to be proven then becomes:$ \ \left(\frac{A + x_n}{n \sqrt[n]{Bx_n}} \right)^n \geq \left(\frac{A}{(n-1)\sqrt[n-1]{B}}...

Re: HSC 2015 4U Marathon \\ $Prove for positive reals$ \ x_1,x_2, \dots , x_n \\ \\ \left(\frac{x_1 + x_2 + \dots + x_n}{n \sqrt[n]{x_1x_2\dots x_n}}\right)^n \geq \left( \frac{x_1 + x_2 + \dots + x_{n-1}}{(n-1) \sqrt[n-1]{x_1 x_2\dots x_{n-1}}} \right)^{n-1}

Another tip I can give, is to use all resources of mathematical knowledge at your disposal. For example, using calculus, one can in fact prove: a^3+ b^3+c^3 \geq 3abc After, all just look at the function f(x) = \frac{1}{3} (x^3 + (b^3 + c^3)) - x (3bc) , \ x > 0 And all you need to do is...

Re: HSC 2015 4U Marathon Better safe than sorry

Re: HSC 2015 4U Marathon Yes, or a more intuitive way of looking at it: x \geq \ln (1+x) \frac{1}{k} \geq \ln \left( 1 + \frac{1}{k} \right) \frac{1}{k} \geq \ln \left(\frac{1+k}{k} \right) = \ln ( k+1) - \ln k \sum_{k=1}^{n} \frac{1}{k} > \ln 2 - \ln 1 + \ln 3 - \ln 2 + \dots + \ln...