HSC 2013-14 MX1 Marathon (archive) (1 Viewer)

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) (1 Viewer)

Well-Known Member

what is that?It is Cowpea

This too shall pass

Member

This too shall pass

This too shall pass

Active Member

This too shall pass

New Member

Member

New Member

New Member

New Member

This too shall pass

Member

This too shall pass

Member

Well-Known Member

This too shall pass

Heroic!

Users Who Are Viewing This Thread (Users: 0, Guests: 1)