# Hard Questions (1 Viewer)

#### leehuan

##### Well-Known Member
Why?
I am an extension 1 student
You're an Ext 1 student and you posted here. Are you saying you want hard Ext 1 maths questions or questions that are actually hard for real mathematicians

##### -insert title here-
Why?
I am an extension 1 student
You're an Ext 1 student and you posted here. Are you saying you want hard Ext 1 maths questions or questions that are actually hard for real mathematicians
Prove or disprove the existence of a universal method to determine whether or not a given equation that defines elliptic curves over ℚ, has finitely, or infinitely many solutions in ℚ.

#### seanieg89

##### Well-Known Member
Prove or disprove the existence of a universal method to determine whether or not a given equation that defines elliptic curves over ℚ, has finitely, or infinitely many solutions in ℚ.
Went to an amazing lecture on this by Venkatesh last year. Fascinating problem.

#### InteGrand

##### Well-Known Member
You're an Ext 1 student and you posted here. Are you saying you want hard Ext 1 maths questions or questions that are actually hard for real mathematicians
I think it was originally posted elsewhere (in Maths Extension 1 forum if I recall correctly), but got moved to this area by a Moderator.

##### -insert title here-
Went to an amazing lecture on this by Venkatesh last year. Fascinating problem.
I know. I've already read up on a similar problem for general diophantine equations and was disappointed (but it didn't go against my expectations) that the answer was a negative.

I won't be surprised if the answer to that problem is a negative though.

#### tywebb

##### dangerman
Here are some more papers from 1956-1962 containing some hard questions:
http://4unitmaths.com/lc1956-1962.pdf
We can see from the 1957 leaving certificate paper that

We can also see from the 2014 HSC Extension 2 exam that

These can be combined to produce a new formula for π in terms of binomial coefficients which was discovered in 2007 by J.C. Toloza:

as follows:

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#### kev@year1223

##### New Member
Hello @tywebb , would like to buy Cambridge worked out solutions for Maths Extension 1 from you. Could you please advise?

#### tywebb

##### dangerman
discovered in 2007 by J.C. Toloza:

$\bg_white \textstyle2+{2\choose2}^{-1}+{3\choose2}^{-1}-{4\choose2}^{-1}-{5\choose2}^{-1}+{6\choose2}^{-1}+{7\choose2}^{-1}-\cdots=\pi$
$\bg_white \text{Toloza's formula essentially relates } \pi \text{ to one diagonal of Pascal's triangle via a pairwise$
$\bg_white \text{alternating sum of reciprocals of binomial coefficients.}$

$\bg_white \text{But Toloza's formula can be generalised to relate }\pi\text{ to every second diagonal of Pascal's triangle.}$

$\bg_white \text{Suppose }T_p:=\textstyle\frac{2p}{2p-1}+\sum_{n=1}^\infty\sum_{r=0}^{2p-1}(-1)^{n-1}{2pn+r\choose 2p}^{-1}\text{ for positive integers }p.$

$\bg_white \text{ Then Toloza's formula is now }T_1=\pi.$

$\bg_white \text{ See if you can prove that for }p>1,$

$\bg_white T_p=\textstyle2\pi\sum_{i=1}^{p-1}(-1)^{i-1}{2p-2\choose i-1}(\csc\frac{\pi i}{2p}-1).$

$\bg_white \text{For example, with } p=2\text{ we find the 4-fold alternating sum }$

$\bg_white \scriptsize{\textstyle\frac{4}{3}+{4\choose4}^{-1}+{5\choose4}^{-1}+{6\choose4}^{-1}+{7\choose4}^{-1}-{8\choose4}^{-1}-{9\choose4}^{-1}-{10\choose4}^{-1}-{11\choose4}^{-1}+\cdots=2(\sqrt2-1)\pi}$

$\bg_white \text{or with }p=3,\text{ the 6-fold alternating sum$

$\bg_white \tiny{\textstyle\frac{6}{5}+{6\choose6}^{-1}+{7\choose6}^{-1}+{8\choose6}^{-1}+{9\choose6}^{-1}+{10\choose6}^{-1}+{11\choose6}^{-1}-{12\choose6}^{-1}-{13\choose6}^{-1}-{14\choose6}^{-1}-{15\choose6}^{-1}-{16\choose6}^{-1}-{17\choose6}^{-1}+\dots=2(5-\frac{8}{3}\sqrt3)\pi}$

$\bg_white \text{etc.}$

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#### tywebb

##### dangerman
$\bg_white \text{Toloza's formula essentially relates } \pi \text{ to one diagonal of Pascal's triangle via a pairwise$
$\bg_white \text{alternating sum of reciprocals of binomial coefficients.}$

$\bg_white \text{But Toloza's formula can be generalised to relate }\pi\text{ to every second diagonal of Pascal's triangle.}$

$\bg_white \text{Suppose }T_p:=\textstyle\frac{2p}{2p-1}+\sum_{n=1}^\infty\sum_{r=0}^{2p-1}(-1)^{n-1}{2pn+r\choose 2p}^{-1}\text{ for positive integers }p.$

$\bg_white \text{ Then Toloza's formula is now }T_1=\pi.$

$\bg_white \text{ See if you can prove that for }p>1,$

$\bg_white T_p=\textstyle2\pi\sum_{i=1}^{p-1}(-1)^{i-1}{2p-2\choose i-1}(\csc\frac{\pi i}{2p}-1).$

$\bg_white \text{For example, with } p=2\text{ we find the 4-fold alternating sum }$

$\bg_white \scriptsize{\textstyle\frac{4}{3}+{4\choose4}^{-1}+{5\choose4}^{-1}+{6\choose4}^{-1}+{7\choose4}^{-1}-{8\choose4}^{-1}-{9\choose4}^{-1}-{10\choose4}^{-1}-{11\choose4}^{-1}+\cdots=2(\sqrt2-1)\pi}$

$\bg_white \text{or with }p=3,\text{ the 6-fold alternating sum$

$\bg_white \tiny{\textstyle\frac{6}{5}+{6\choose6}^{-1}+{7\choose6}^{-1}+{8\choose6}^{-1}+{9\choose6}^{-1}+{10\choose6}^{-1}+{11\choose6}^{-1}-{12\choose6}^{-1}-{13\choose6}^{-1}-{14\choose6}^{-1}-{15\choose6}^{-1}-{16\choose6}^{-1}-{17\choose6}^{-1}+\dots=2(5-\frac{8}{3}\sqrt3)\pi}$

$\bg_white \text{etc.}$
I found a proof which is attached. There might be an easier way to do it, or at least a simplification of the attached proof.

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