HSC 2013-14 MX1 Marathon (archive) (2 Viewers)

hayabusaboston · Aug 2, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Considering the identity:

$(1+x)^{n+m} = (1+x)^n (1+x)^m$

Then expanding both sides, equate co-efficient of $x^k$ on both sides knowing that $m>n>k$ , the result comes immediately.

=======

$$Prove$ \ \ \frac{1}{2}\binom{n}{0} - \frac{1}{3} \binom{n}{1} + \frac{1}{4} \binom{n}{2} - \dots + \frac{(-1)^n}{n+2} \binom{n}{n} = \frac{1}{(n+1)(n+2)}$

Consider (1+x)^n=nC0+nC1x+...nCnx^n, integrate twice, find ur C's, and let x=-1?
From first glance dats what it looks like.

RealiseNothing · Aug 2, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Considering the identity:

$(1+x)^{n+m} = (1+x)^n (1+x)^m$

Then expanding both sides, equate co-efficient of $x^k$ on both sides knowing that $m>n>k$ , the result comes immediately.

Damn I forgot you could do it that way. I had a combinations method in mind.

Sy123 · Aug 2, 2013

Re: HSC 2013 3U Marathon Thread

hayabusaboston said:
Consider (1+x)^n=nC0+nC1x+...nCnx^n, integrate twice, find ur C's, and let x=-1?
From first glance dats what it looks like.

You will get on the denominators 2*3, 3*4, 4*5, .... , (n+1)(n+2) while for this one the denominators are only 2, 3, 4, .... , n+2

JJ345 · Aug 2, 2013

Re: HSC 2013 3U Marathon Thread

Hmm...this question looks familiar.
Is it something like using the binomial expansion for (1+x)^n, then multiplying both sides by x, integrating both sides (the LHS should have an expression that's slightly harder to integrate than usual), sub in x=0 to find the value of the constant of integration, then finally sub. x=-1. Not too sure, haven't done this in a while.

Sy123 · Aug 2, 2013

Re: HSC 2013 3U Marathon Thread

JJ345 said:
Hmm...this question looks familiar.
Is it something like using the binomial expansion for (1+x)^n, then multiplying both sides by x, integrating both sides (the LHS should have an expression that's slightly harder to integrate than usual), sub in x=0 to find the value of the constant of integration, then finally sub. x=-1. Not too sure, haven't done this in a while.

A little different to what I had in mind, integrating x(1-x)^n from 0 to 1
But essentially the same thing

=====

Here is question I made:

$In this question take$

$\phi = \frac{1+\sqrt{5}}{2} , \ \tau = \frac{1-\sqrt{5}}{2}$

$Consider the identity$

$(x-\phi)(x-\tau) = x^2-x-1$

$Prove that$

$\sum_{k=0}^{2n} \binom{2n}{k}^2 \phi^{2n-k} \tau^k = \sum_{k=0}^n \binom{2n}{k} \binom{2n-k}{k}(-1)^k$

Sy123 · Aug 4, 2013

Re: HSC 2013 3U Marathon Thread

$$A particle is moving at a height of$ \ b \ $above the ground under conditions of gravity$ \ g$

$$The particle is oscillating in simple harmonic motion about $ \ y= 0 \ $parallel to the ground with an amplitude of$ \ b \ $and a period of$ \ \frac{2\pi}{n}$

$At some time the particle is released and the particle undergoes projectile motion$

$Show that the maximum horizontal range that can be achieved is$

$b\sqrt{\frac{2bn^2+g}{g}}$

dunjaaa · Aug 5, 2013

Re: HSC 2013 3U Marathon Thread

Is there meant to be a (+ g) in the inside? I got the expression without the +g

Sy123 · Aug 5, 2013

Re: HSC 2013 3U Marathon Thread

dunjaaa said:
Is there meant to be a (+ g) in the inside?

I don't think so, I could be wrong but I'm getting what the question sasys.

seanle · Aug 7, 2013

Re: HSC 2013 3U Marathon Thread

hi 2x^2 + ax + b + 3 =0 HAs real roots. Find the miniumum value of a^2 + b^2
Tried this but could not get the right answer. A cambridge extension question from year 11 book for 3 unit

chris_power_96 · Aug 7, 2013

Re: HSC 2013 3U Marathon Thread

seanle said:
hi 2x^2 + ax + b + 3 =0 HAs real roots. Find the miniumum value of a^2 + b^2
Tried this but could not get the right answer. A cambridge extension question from year 11 book for 3 unit

I dont think this is right but ill have a shot anyway.
So you find a and b through sum and product of roots then add the square of both them which should get you
a^2+b^2=4c^2+4d^2+8cd+4c^2d^2-12cd+9 (c and d are the roots of the equation)
Now for a and b to me minimum, c and d approach 0, meaning the min is 9

seanle · Aug 7, 2013

Re: HSC 2013 3U Marathon Thread

hi i tried using the discriminant let it equal zero for a minimum but i got 8 for an answer.
thank for your answer.

seanle · Aug 7, 2013

Re: HSC 2013 3U Marathon Thread

a^2-4.2(b+3) is greater than or equal to zero

for a minimum a^2 = 8b + 24

let a^2 + b^2 = 8b + 24 + b^2 for b+3 = 0 this does give
the answer as 9 but why should i let b+ 3 =0 ?
is it a critical point?

seanle · Aug 7, 2013

Re: HSC 2013 3U Marathon Thread

a = the square root of 8b + 24
for real roots b+3 is more than or equal to zero. So the minimum value of b is -3.

Sy123 · Aug 9, 2013

Re: HSC 2013 3U Marathon Thread

$$Prove$ \ \binom{n}{0} + \frac{1}{3} \binom{n}{2} + \frac{1}{5} \binom{n}{4} + \frac{1}{7} \binom{n}{6} + \dots = \frac{2^{n-1}}{n+1}$

JJ345 · Aug 9, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$Prove$ \ \binom{n}{0} + \frac{1}{3} \binom{n}{2} + \frac{1}{5} \binom{n}{4} + \frac{1}{7} \binom{n}{6} + \dots = \frac{2^{n-1}}{n+1}$

I'm thinking of expanding (1+x)^n, integrating, sub x=0 to find c, then let x=1 call this (1)
Repeat the process but in the final step sub x=-1 call this (2)
Then (1)-(2) will yield you a similar result---> But with 2^n on the numerator on the LHS???

Sy123 · Aug 9, 2013

Re: HSC 2013 3U Marathon Thread

JJ345 said:
I'm thinking of expanding (1+x)^n, integrating, sub x=0 to find c, then let x=1 call this (1)
Repeat the process but in the final step sub x=-1 call this (2)
Then (1)-(2) will yield you a similar result---> But with 2^n on the numerator on the LHS???

Yep the steps are correct, however when you do the (1) - (2) we get 2 times the sum on the LHS = 2^n/ (n+1) dividing by 2 gives 2^(n-1)

====

$Prove$

$\binom{2n}{0}^2 - \binom{2n}{1}^2 + \binom{2n}{2}^2 - \binom{2n}{3}^2 + \dots - \binom{2n}{2n-1}^2 + \binom{2n}{2n}^2 = (-1)^n \binom{2n}{n}$

JJ345 · Aug 9, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Yep the steps are correct, however when you do the (1) - (2) we get 2 times the sum on the LHS = 2^n/ (n+1) dividing by 2 gives 2^(n-1)

====

$Prove$

$\binom{2n}{0}^2 - \binom{2n}{1}^2 + \binom{2n}{2}^2 - \binom{2n}{3}^2 + \dots - \binom{2n}{2n-1}^2 + \binom{2n}{2n}^2 = (-1)^n \binom{2n}{n}$

I think i'm missing something here, but how do we get this?
When we integrated didn't the LHS become [(1+x)^n+1/n+1] +c ?

seanieg89 · Aug 9, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Yep the steps are correct, however when you do the (1) - (2) we get 2 times the sum on the LHS = 2^n/ (n+1) dividing by 2 gives 2^(n-1)

====

$Prove$

$\binom{2n}{0}^2 - \binom{2n}{1}^2 + \binom{2n}{2}^2 - \binom{2n}{3}^2 + \dots - \binom{2n}{2n-1}^2 + \binom{2n}{2n}^2 = (-1)^n \binom{2n}{n}$

I am quite sure the numerator should be 2^n.

Edit: In the previous question that is.

Sy123 · Aug 9, 2013

Re: HSC 2013 3U Marathon Thread

JJ345 said:
I think i'm missing something here, but how do we get this?
When we integrated didn't the LHS become [(1+x)^n+1/n+1] +c ?

When integrating (1+x)^n, we get $\frac{(1+x)^{n+1}}{n+1} + c = x \times ($Something$)$

Then x=0 to find c= - 1/(n+1)

Then, $\frac{(1+x)^n - 1}{n+1} = \frac{n}{0}x + \frac{1}{2} \binom{n}{1} x^2 + \dots + \binom{n}{n} \frac{1}{n+1}$

EDIT: Yep I'm wrong I made a mistake, apologies

HeroicPandas said:
$\rightarrow \\\\ (1+x)^{2n}(1-x)^{2n} \equiv (1-x^2)^{2n}\\\\(1+x)^{2n} = \binom{2n}{0} + \binom{2n}{1}x +\binom{2n}{2}x^2 + \dots + \binom{2n}{2n}x^{2n} \dots [1] \\ (1-x)^{2n} = \binom{2n}{0} - \binom{2n}{1}x +\binom{2n}{2}x^2 - \dots + (-1)^{2n}\binom{2n}{2n}x^{2n} \dots [2] \\ \\ (1-x^2)^{2n} = \binom{2n}{0} -\binom{2n}{1}x^2 + \binom{2n}{2}x^4 - \dots (-1)^{2n}x^{2n} \dots [3] \\ \\$[1] \cdot [2] = [3]\\\\ $ $Equate coefficient of $x^{2n} $ and use $\binom{n}{r} = \binom{n}{n-r} \\ \\$

Yep well done

HeroicPandas · Aug 9, 2013

Re: HSC 2013 3U Marathon Thread

I found a mistake in my working out for the RHS $(-1)^n \binom{2n}{n}$

If i equate coefficients of x^(2n) for [3], i get $\binom{2n}{2n}$

HSC 2013-14 MX1 Marathon (archive) (2 Viewers)

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