HSC Mathematics Marathon (1 Viewer)

hup · Oct 16, 2011

show that

$\frac{1}{n!}<\frac{1}{2^{n-1}}$

for all $n\geq 3$

without the use of induction

bobthebuilder94 · Oct 16, 2011

i was gonna do an awesome proof where i differentiate it and then do some amazing stuff....but then realised my maths is too limited

can't differentiate 1/x!

slyhunter · Oct 16, 2011

bobthebuilder94 said:
i was gonna do an awesome proof where i differentiate it and then do some amazing stuff....but then realised my maths is too limited can't differentiate 1/x!

$-\frac{1}{x^2}$

?

AAEldar · Oct 16, 2011

slyhunter said:
$-\frac{1}{x^2}$

?

He's talking about the factorial, which there is no derivative of.

mitchy_boy · Oct 16, 2011

hup said:
show that

$\frac{1}{n!}<\frac{1}{2^{n-1}}$

for all $n\geq 3$

without the use of induction

n! grows faster than $x^n$
assuming constant x.

therefore true

this is a troll question

mitchy_boy · Oct 16, 2011

Find the coordinates of the focus of the parabola $12y = x^2 -6x - 3$ .

ohexploitable · Oct 16, 2011

mitchy_boy said:
Find the coordinates of the focus of the parabola $12y = x^2 -6x - 3$ .

srsly?

mitchy_boy · Oct 16, 2011

ohexploitable said:
srsly?

This is 2unit m8, go back to your gen maffs.

mitchy_boy · Oct 16, 2011

it was the first question i found in a book on my floor

hup · Oct 16, 2011

mitchy_boy said:
n! grows faster than $x^n$
assuming constant x.

therefore true

this is a troll question

its doable

cutemouse · Oct 17, 2011

hup said:
show that

$\frac{1}{n!}<\frac{1}{2^{n-1}}$

for all $n\geq 3$

without the use of induction

1/n! = 1/(n(n-1)(n-2)...(3)(2)1) < 1/(2*2*2*...*2*2*1) = 1/2^(n-1) for n>=3. QED

hup · Oct 17, 2011

cutemouse said:
1/n! = 1/(n(n-1)(n-2)...(3)(2)1) < 1/(2*2*2*...*2*2*1) = 1/2^(n-1) for n>=3. QED

whats qed

slyhunter · Oct 17, 2011

QED: quod erat demonstrandum

Latin for "which was to be demonstrated". Usually placed at the conclusion of a mathematical proof.

hup · Oct 17, 2011

apollo1 said:
wtf did u do. i dont follow

list out

3>2
4>2
...
n>2

multiply

3x4x...xn>2^(n-2)
1x2x3x....xn>2^(n-1)
he just did it in one step

Trebla · Oct 17, 2011

Trebla said:
Suppose that

$z = \cos\theta + i\sin\theta$

Use the results

$z^n+z^{-n}=2\cos n\theta$

and

$z^n-z^{-n}=2i\sin n\theta$

to show that

$\displaystyle\sum_{k=0}^{n}\cos[(2k+1)\theta]=\dfrac{\sin[2(n+1)\theta]}{2\sin\theta}$

.

seanieg89 · Oct 18, 2011

RANK 1 · Oct 18, 2011

u're supposed to give a new question after answering one.....

apollo1 · Oct 18, 2011

ill post one on behalf of seanieg

Screen shot 2011-10-18 at 2.00.52 PM.png

AAEldar · Oct 18, 2011

apollo1 said:
ill post one on behalf of seanieg
View attachment 23329

$\sum_{r=1}^{r=n} \frac{r\binom{n}{r}}{\binom{n}{r-1}}=\frac{n}{2}(n+1) \\\\ LHS=\sum_{r=1}^{r=n} \frac{rn!}{(n-r)!r!}\div \frac{n!}{(n-r+1)!(r-1)!} \\\\ =\sum_{r=1}^{r=n} \frac{n!}{(n-r)!(r-1)!}\times \frac{(n-r+1)!(r-1)!}{n!} \\\\ =\sum_{r=1}^{r=n} n-r+1 \\\\ =\frac{n}{2}(n+1) \\\\ =RHS$

I don't have another question though... I'll try find one somewhere >.<

seanieg89 · Oct 19, 2011

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