STEP Prep Thread 2017

Hello everyone. I am probably being really thick here but on the STEP II 1987 specimen question 16 with the tennis tournament I can see why there are 1/2^n -1 different pairings for the first round but I can't see why (2^n -2)C2/2^nC2 doesn't work because there are 2^nC2 ways of choosing pairs in total and (2^n -2)C2 in which any particular pair is chosen because there are 2^n -2 unpaired people surely???

Can you DM me the STEP mechanics guide when you're done with it? It sounds amazing tbh!!!

It's linked in post 228 of this thread :-)

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Hi guys,*

I've been on-and-off working on some notes for step mechanics, on the basis that it's often a little underrepresented hereabouts. It's by no means finished, and probably full of errors, but I'll put it up now so you can suggest changes and hopefully benefit from it.

K

https://docs.google.com/document/d/1OiG2tmZ1OENPK9U2ob8HUDyzbBzsdRvFdCANbH4a17Y/edit?usp=sharing*

In case others miss this, quoting again. Zacken: is this in the OP?

4x^2 +16xy +y^2 +24x =0. Prove that x is less than or equal to zero and greater than or equal to 2/5, and find similar restrictions for y. I started by rearranging the above equation to (2x+y)^2 +12x(y+2), and then (2x+y)^2 =-12x(y+2), so since LHS is positive, then RHS must be positive, but that can only happen if x is less than or equal to zero, (to cancel with the negative 12), or if (y+2) is less than or equal to zero, or y is less than or equal to-2. So I've proved one of the inequalities given. But how do I prove the other one.

4x^2 +16xy +y^2 +24x =0. Prove that x is less than or equal to zero and greater than or equal to 2/5, and find similar restrictions for y. I started by rearranging the above equation to (2x+y)^2 +12x(y+2), and then (2x+y)^2 =-12x(y+2), so since LHS is positive, then RHS must be positive, but that can only happen if x is less than or equal to zero, (to cancel with the negative 12), or if (y+2) is less than or equal to zero, or y is less than or equal to-2. So I've proved one of the inequalities given. But how do I prove the other one.

Whist your initial approach is along the right lines (with completing the square), the way in which you do it makes life rather difficult for you at the end. In general, if possible, it's best to avoid mixing up your $x$'s and $y$'s together, you'd much rather have them separate, especially when trying to form inequalities with them.

To that end, you'd be best off completing the square as $(y+8x)^2 = 60x^2 + 24x$ then as you said, squares are always non-negative, so you get an easy quadratic inequality for $x$ that you can solve easily, giving you the requisite bounds.

With that in mind, I'll let you now (re-)complete the square such a way that you get $[g(x,y)]^2=h(y)$ so that you can solve the (hopefully quadratic) inequality $h(y) \geq 0$

Whist your initial approach is along the right lines (with completing the square), the way in which you do it makes life rather difficult for you at the end. In general, if possible, it's best to avoid mixing up your $x$'s and $y$'s together, you'd much rather have them separate, especially when trying to form inequalities with them.

To that end, you'd be best off completing the square as $(y+8x)^2 = 60x^2 + 24x$ then as you said, squares are always non-negative, so you get an easy quadratic inequality for $x$ that you can solve easily, giving you the requisite bounds.

With that in mind, I'll let you now (re-)complete the square such a way that you get $[g(x,y)]^2=h(y)$ so that you can solve the (hopefully quadratic) inequality $h(y) \geq 0$

It should be $(y+8x)^2 = 60x^2 - 24x$ which yields $12x(5x-2)>0$ as required.

This question asks you to find the coefficient of x^n of the equation 9/[(2-x)^2 (1+x)]. This equation will become 9[(2-x)^-2 (1+x)^-1]. Expanding the first bracket, the coefficient of the x^n term is {[(-2)(-3)...(2)(1)]/n!} *([1/2]x)^n, and for the second bracket it is {[(-1)(-2)...(2)(1)]/n!} * (x)^n. How do I work out the multiple of those two numbers?

Well that's not how you find the coefficient of x^n in the expansion. When you multiply those things together you're getting a term in x^2n which doesn't help you. For each k from 0 to n you want to pair off each term in x^k in one of the expansions with each term in x^(n-k) in the other.
edit: also those expansions don't make sense. You would have in the first case (-2)(-3)...(-2-n+1) rather than what you have, similarly (-1)(-2)...(-n) in the second case. And there seem to be a few other technical errors...check your working carefully in boring slogs like this

They're asking for the coefficient of x^n, so why do I need to find the term x^n-k for one of them? Why will I get the x^2n term in my expansion, I only used the the term x^n for each expansion. Btw, I'm pretty sure my expansions are correct. To remind you the expansion formula is n(n-1)(n-2)...
(n-(n-1))/n! *(a)^n, where the expanded bracket is in the form (1+a)^n.

What question is this? So I am clear on precisely what it is asking. Yes you have found the x^n term in those two different series, but the expansion of the entire thing involves multiplying two expansions together. Multiplying x^n term with an x^n term gives you an x^2n term..
no that expansion doesn't work in these cases with fractional or negative powers. to get the kth term where the power being raised to is n you want (n)(n-1)...(n - k + 1)

Specimen 86 s1

So basically, you're saying find the kth term for one of the brackets, and the (n-k)th term for the other, and multiply them?