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HSC 2013-14 MX1 Marathon (archive) (6 Viewers)

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SpiralFlex

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Re: HSC 2013 3U Marathon Thread

Didn't see that first time, didn't edit it in quick enough
 

kev-

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Re: HSC 2013 3U Marathon Thread

For the curve y=f(x), it is given that f'(x)=sin^2 x and the curve passes through the point (pi/4, pi/8). Find the equation of the curve.
 

Sy123

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Re: HSC 2013 3U Marathon Thread




(yes it is very easy to prove without induction, but this question isn't meant for that)
 
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RealiseNothing

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Re: HSC 2013 3U Marathon Thread




(yes it is very easy to prove without induction, but this question isn't meant for that)
Since 'n' is even, let , then we have:



Now if this is divisible by 12 like the question claims, then all we have to do is divide by 12 and show that the result is an integer for



Grouping together the fractions gives:



Since 'k' is an integer, then is an integer, and we can disregard it. If we prove the 2nd half is an integer, then the solution is complete, so consider:



When 'k 'is a multiple of 3, then it is an integer, so we only have to prove it is an integer for non-multiples of 3.

Consider what is, it can be expressed as the sum of the odd integers, so multiplying it by two would give:



Now after this we add 1 since it is . Notice the difference in the series 2,6,10,14,etc is 4. So the difference is 1 more than a multiple of 3. Since the first term is 2 (one less than a multiple of 3) then we add 1, it becomes a multiple of 3. The next number, since we add 4, also becomes a multiple of 3. This is because:

and hence it also a multiple of 3. Thus in if 'k' is one more than multiple of 3, then next number after it is also a multiple of 3. Hence all that is left to prove is that all integers such that allow to be divisible by 3.

Now it works for quite trivially. Considering the common difference of the odd numbers when multipled by two is 4, adding 4 three times in total adds 12. Now 12 is divisible by 3 and hence all integers of 'k' such that must allow to be divisible by 3. Which concludes the proof.
 
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Sy123

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Re: HSC 2013 3U Marathon Thread

Since 'n' is even, let , then we have:



Now if this is divisible by 12 like the question claims, then all we have to do is divide by 12 and show that the result is an integer for



Grouping together the fractions gives:



Since 'k' is an integer, then is an integer, and we can disregard it. If we prove the 2nd half is an integer, then the solution is complete, so consider:



When 'k 'is a multiple of 3, then it is an integer, so we only have to prove it is an integer for non-multiples of 3.

Consider what is, it can be expressed as the sum of the odd integers, so multiplying it by two would give:



Now after this we add 1 since it is . Notice the difference in the series 2,6,10,14,etc is 4. So the difference is 1 more than a multiple of 3. Since the first term is 2 (one less than a multiple of 3) then we add 1, it becomes a multiple of 3. The next number, since we add 4, also becomes a multiple of 3. This is because:

and hence it also a multiple of 3. Thus in if 'k' is one more than multiple of 3, then next number after it is also a multiple of 3. Hence all that is left to prove is that all integers such that allow to be divisible by 3.

Now it works for quite trivially. Considering the common difference is 4, adding 4 three times in total adds 12. Now 12 is divisible by 3 and hence all integers of 'k' such that must allow to be divisible by 3. Which concludes the proof.
Nice work!

I was aiming for someone to undergo the induction step:

Step 3: n = k + 2

And if our initial condition is: Step 1: n= 2, then if we modify Step 3 to be such above then we can prove for all even integers.
Similarly to prove something for all odd integers, we only need to configure Step 1 to be n=1.
 

Sy123

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Re: HSC 2013 3U Marathon Thread





 

Sy123

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Re: HSC 2013 3U Marathon Thread

The easier way, and I will allow it, is to consider (1+x)^n = P(x)
Then consider the sum of roots of P(x) 1 at a time
2 at a time
3 at a time

and so on....
(all the roots are -1)
 

omgiloverice

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Re: HSC 2013 3U Marathon Thread

This thread is as dead as a roadside pigeon ): ...

1. Prove by mathematical induction that for all integers that is less than
2. Prove by mathematical induction that for all integers

I'm posting these because I haven't got a satisfactory answer yet.
 
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braintic

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Re: HSC 2013 3U Marathon Thread

You mean those pigeons I aim at with my car?
 
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Re: HSC 2013 3U Marathon Thread

This thread is as dead as a roadside pigeon ): ...

1. Prove by mathematical induction that for all integers that is less than
2. Prove by mathematical induction that for all integers

I'm posting these because I haven't got a satisfactory answer yet.
You mean those pigeons I aim at with my car?
oi
 

Sy123

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Re: HSC 2013 3U Marathon Thread

Here is my own little proof of the 'famous' identity (its probably similar to others, but I haven't seen any)



 

RealiseNothing

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Re: HSC 2013 3U Marathon Thread

:|

I have a real bad memory.
lol

I'll start people off with the idea of how to approach the question, so some one else can attempt it (I assume it scared people off since it "looks" hard)

Consider the binomial expansion of by itself, then consider the binomial expansion of

Then equate the co-efficients of the RHS and LHS of and go from there.

(this is how I would do it, not sure if you have something else in mind).
 
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