HSC 2013 MX2 Marathon (archive) (2 Viewers)

Sy123 · Aug 14, 2013

Re: HSC 2013 4U Marathon

A good Putnam problem that can be done with elementary knowledge

$Evaluate$

$\prod_{n=2}^{\infty} \frac{n^3-1}{n^3+1} \equiv \left( \frac{2^3-1}{2^3+1} \right ) \left ( \frac{3^3-1}{3^3+1}\right ) \left(\frac{4^3-1}{4^3+1} \right ) \left(\frac{5^3-1}{5^3 + 1} \right ) \dots \dots$

Sy123 · Aug 15, 2013

Re: HSC 2013 4U Marathon

$Find$

$\sum_{k=1}^n \frac{k}{(k+1)!}$

JJ345 · Aug 15, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Find$

$\sum_{k=1}^n \frac{k}{(k+1)!}$

Is the answer 1-1/n!-n/(n+1)! ?

Sy123 · Aug 15, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Is the answer 1-1/n!-n/(n+1)! ?

I got: 1 - 1/n! + n/(n+1)! (only after a manipulation of my answer, I did not get it directly).

What method did you use?

Drongoski · Aug 15, 2013

Re: HSC 2013 4U Marathon

$\frac {k}{(k+1)!} = \frac {k+1}{(k+1)!} - \frac {1}{k+1()!} = \frac {1}{k!} - \frac {1}{(k+1)!} \\ \\ \therefore \sum _1 ^n \frac {k}{(k+1)!} = (\frac {1}{1!} - \frac {1}{2!}) + (\frac {1}{2!} - \frac {1}{3!}) + \dots + (\frac {1}{n!} - \frac {1}{(n+1)!}) \\ \\ = \frac {1}{1!} - \frac {1}{(n+1)!} = 1 - \frac {1}{(n+1)!}$

Sy123 · Aug 15, 2013

Re: HSC 2013 4U Marathon

Drongoski said:
$\frac {k}{(k+1)!} = \frac {k+1}{(k+1)!} - \frac {1}{k+1()!} = \frac {1}{k!} - \frac {1}{(k+1)!} \\ \\ \therefore \sum _1 ^n$

Yep this was my method

seanieg89 · Aug 15, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Is the answer 1-1/n!-n/(n+1)! ?

A quick (though crude) way to check your answers to things like this is chuck in n=1.

JJ345 · Aug 15, 2013

Re: HSC 2013 4U Marathon

Yes, sorry I made a mistake when typing it out it was meant to be 1 - 1/n! + n/(n+1)!

seanieg89 · Aug 16, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
A good Putnam problem that can be done with elementary knowledge

$Evaluate$

$\prod_{n=2}^{\infty} \frac{n^3-1}{n^3+1} \equiv \left( \frac{2^3-1}{2^3+1} \right ) \left ( \frac{3^3-1}{3^3+1}\right ) \left(\frac{4^3-1}{4^3+1} \right ) \left(\frac{5^3-1}{5^3 + 1} \right ) \dots \dots$

No-one else seems to be answering it, so:

$P_N=\prod_{n=2}^N \frac{n-1}{n+1} \cdot \prod_{n=2}^N \frac{(n+1/2)^2+3/4}{(n-1/2)^2+3/4}=\frac{2}{N(N+1)}\cdot \frac{(N+1/2)^2+3/4}{3}.$

by telescoping the two products separately. As N tends to infinity, this expression tends to 2/3 (divide numerator and denominator by N^2 and all "non-leading" terms will die.)

So the infinite product converges to 2/3.

shongaponga · Aug 17, 2013

Re: HSC 2013 4U Marathon

Slightly advanced for 3U students + 3U marathon forum seems to be dead so i'll just post it here:

$\\ $ Prove by Mathematical Induction that for all positive values of $ n, \\\\ \tan^{-1}(\frac{1}{2}) + \tan^{-1}(\frac{1}{8}) + \tan^{-1}(\frac{1}{18}) +...+ \tan^{-1}(\frac{1}{2n^2}) = \frac{\pi}{4} - \tan^{-1}(\frac{1}{2n+1})$

Sy123 · Aug 17, 2013

Re: HSC 2013 4U Marathon

shongaponga said:
Slightly advanced for 3U students + 3U marathon forum seems to be dead so i'll just post it here:

$\\ $ Prove by Mathematical Induction that for all positive values of $ n, \\\\ \tan^{-1}(\frac{1}{2}) + \tan^{-1}(\frac{1}{8}) + \tan^{-1}(\frac{1}{18}) +...+ \tan^{-1}(\frac{1}{2n^2}) = \frac{\pi}{4} - \tan^{-1}(\frac{1}{2n+1})$

$$We want to find a sequence$ \ u_k$

$$Such that$ \ \ \tan^{-1}(u_{k-1}) - \tan^{-1}(u_k) = \tan^{-1} \frac{1}{2k^2}$

From inspection, $u_0 = 1$
$u_n = \frac{1}{2n+1}$

$\therefore \ u_k = \frac{1}{2k+1} \ \ u_{k-1} = \frac{1}{2k-1}$

$$We verify this by considering the identity$ \ \ \tan^{-1} A - \tan^{-1}B = \tan^{-1} \left(\frac{A-B}{1+AB} \right)$ and substituting in appropriately.

Thus

$\sum_{k=1}^n \tan^{-1} \frac{1}{2n^2} = \sum_{k=1}^n \tan^{-1} \frac{1}{2k-1} - \tan^{-1} \frac{1}{2k+1} = \tan^{-1}(1) - \tan^{-1} \frac{1}{2n+1} = \frac{\pi}{4} - \tan^{-1} \left(\frac{1}{2n+1} \right )$

=================

$$Without using a calculator at all, how many digits are in the number$ \ 2^{1000} ?$

$Your answer may be wrong within$ \ \pm 50 \ \ $of the correct answer$

chris_power_96 · Aug 17, 2013

Re: HSC 2013 4U Marathon

The angles of a triangle are such that the largest is one right angle in excess of the smallest. Given that the lengths of the sides of the triangle form an arithmetic sequence, find the ratios of the side lengths

JJ345 · Aug 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$We want to find a sequence$ \ u_k$

$$Such that$ \ \ \tan^{-1}(u_{k-1}) - \tan^{-1}(u_k) = \tan^{-1} \frac{1}{2k^2}$

From inspection, $u_0 = 1$
$u_n = \frac{1}{2n+1}$

$\therefore \ u_k = \frac{1}{2k+1} \ \ u_{k-1} = \frac{1}{2k-1}$

$$We verify this by considering the identity$ \ \ \tan^{-1} A - \tan^{-1}B = \tan^{-1} \left(\frac{A-B}{1+AB} \right)$ and substituting in appropriately.

Thus

$\sum_{k=1}^n \tan^{-1} \frac{1}{2n^2} = \sum_{k=1}^n \tan^{-1} \frac{1}{2k-1} - \tan^{-1} \frac{1}{2k+1} = \tan^{-1}(1) - \tan^{-1} \frac{1}{2n+1} = \frac{\pi}{4} - \tan^{-1} \left(\frac{1}{2n+1} \right )$

=================

$$Without using a calculator at all, how many digits are in the number$ \ 2^{1000} ?$

$Your answer may be wrong within$ \ \pm 50 \ \ $of the correct answer$

If not using the calculator at all, the only logic I can think of is--> we know 2^10 (1024) is slightly greater than 10^3
So (2^10)^100 should be slightly greater than (10^3)^100=10^300--> So it should have slightly more than 300 digits, I would estimate 310(is that in give or take 50 range?)

I'm thinking of finding it through using logs--> But i would probably need a calculator at some stage for that.

Drongoski · Aug 17, 2013

Re: HSC 2013 4U Marathon

Is it 301 or 302?

Sy123 · Aug 17, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
If not using the calculator at all, the only logic I can think of is--> we know 2^10 (1024) is slightly greater than 10^3
So (2^10)^100 should be slightly greater than (10^3)^100=10^300--> So it should have slightly more than 300 digits, I would estimate 310(is that in give or take 50 range?)

Yep that is quite accurate of an estimate, the correct value is 302

Its better than my method though which was: (10^(log(2) * 1000) ), (using a calculator we get immediately that log(2) = 0.3016....) but my intention was for the answer to contain an attempt using integration to rationally approximate the value of log(2).
I was able to get like 340 using that lol

EDIT: It should be 302 since, 10^1 has 2 digits, 10^3 has 4 digits, thus 10^(301) should have 302 digits.

Drongoski · Aug 17, 2013

Re: HSC 2013 4U Marathon

log₁₀2 = 0.3010 ....

.: log₁₀ 2¹⁰⁰⁰ = 301.02999 ..

.: answer = 301 + 1

(301 is the whole number part before the mantissa of the log of the number and is 1 less than the number of digits of the number excluding the fractional part)

Sy123 · Aug 18, 2013

Re: HSC 2013 4U Marathon

$Prove$

$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots < 3$

RealiseNothing · Aug 18, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove$

$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots < 3$

$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots < 1 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = 3$

Sy123 · Aug 18, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots < 1 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = 3$

Yep well done

$$Solve for all$ \ \ x \ $and$ \ y$

$\frac{1}{x} + \frac{1}{y} = - 1$

$x^3+y^3 = 4$

seanieg89 · Aug 18, 2013

Re: HSC 2013 4U Marathon

In a class consisting of n mathematics students, every student belongs to exactly one friendship group and each friendship group has at least 2 students.

Let S(n) be the number of possible arrangements of friendship groups.

Find a recursion relation for S(n) and calculate S(n) for n=1 to 6.

HSC 2013 MX2 Marathon (archive) (2 Viewers)

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