HSC 2013-14 MX1 Marathon (archive) (8 Viewers)

asianese · Feb 24, 2013

Re: HSC 2013 3U Marathon Thread

True that.

Sy123 · Mar 2, 2013

Re: HSC 2013 3U Marathon Thread

$Simplify$

$\binom{2n}{0} + \binom{2n}{2} + \binom{2n}{4} + \dots + \binom{2n}{2n}$

RealiseNothing · Mar 2, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Simplify$

$\binom{2n}{0} + \binom{2n}{2} + \binom{2n}{4} + \dots + \binom{2n}{2n}$

Considering Pascal's triangle, we know that the sum of two adjacent numbers is the number directly below them. Hence we can write the question as the sum of the numbers in the row above it:

$\binom{2n-1}{0} + \binom{2n-1}{1} + \binom{2n-1}{2} + \binom{2n-1}{3} + \binom{2n-1}{4} + \dots + \binom{2n-1}{2n-1}$

This is just the sum of the $2n-1th$ row of Pascal's triangle, which is $2^{2n-1}$

Sy123 · Mar 2, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
Considering Pascal's triangle, we know that the sum of two adjacent numbers is the number directly below them. Hence we can write the question as the sum of the numbers in the row above it:

$\binom{2n-1}{0} + \binom{2n-1}{1} + \binom{2n-1}{2} + \binom{2n-1}{3} + \binom{2n-1}{4} + \dots + \binom{2n-1}{2n-1}$

This is just the sum of the $2n-1th$ row of Pascal's triangle, which is $2^{2n-1}$

Very elegant, nice work, my way was a little longer by considering expansions of (1+x)^2n and (1-x)^2n

Sy123 · Mar 9, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 · Mar 11, 2013

Re: HSC 2013 3U Marathon Thread

A nice result I've proven

$Prove using Binomial Theorem or Combinatorics$

$\binom{n}{n} + \binom{n+1}{n} + \binom{n+2}{n} + \dots + \binom{2n}{n} = \binom{2n}{n+1}$

seanieg89 · Mar 11, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
A nice result I've proven

$Prove using Binomial Theorem or Combinatorics$

$\binom{n}{n} + \binom{n+1}{n} + \binom{n+2}{n} + \dots + \binom{2n}{n} = \binom{2n}{n+1}$

Err, not true. The last term on the LHS alone is greater than the RHS...

Sy123 · Mar 11, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
Err, not true. The last term on the LHS alone is greater than the RHS...

I considered the geometric series:

$(1+x)^n + (1+x)^{n+1} + \dots + (1+x)^{2n} = \frac{(1+x)^n ((1+x)^n -1)}{(1+x)-1} = \frac{(1+x)^{2n}-(1+x)^n}{x}$

Equate co-efficient of x^n on both sides of equation. Anything illegal in the proof?

seanieg89 · Mar 11, 2013

Re: HSC 2013 3U Marathon Thread

Yes. Check the number of terms on the LHS, there are n+1, not n. A minor change needs to be made to your formula for the sum of a finite geometric series.

Sy123 · Mar 11, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
Yes. Check the number of terms on the LHS, there are n+1, not n. A minor change needs to be made to your formula for the sum of a finite geometric series.

:/ Can't believe I missed that, thanks for the heads up. I'll ask this question again once everyone has forgotten about it.

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

New result:

$Prove using the Binomial Theorem or Combinatorics$

$\frac{1}{2} \binom{n}{0} + \frac{1}{3} \binom{n}{1} + \dots + \frac{1}{n+2} \binom{n}{n} = \frac{ 2^{n+1} \cdot n + 1}{(n+1)(n+2)}$

Sy123 · Mar 19, 2013

Re: HSC 2013 3U Marathon Thread

$$Find$ \ \ \ 100^2+99^2-98^2-97^2+96^2+95^2-94^2-93^2+ \dots + 4^2 + 3^2 -2^2 -1^2$

(The signs alternate in pairs)
(Provide proof of your value, either with latex or a brief outline of steps taken)

Sy123 · Mar 19, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
:/ Can't believe I missed that, thanks for the heads up. I'll ask this question again once everyone has forgotten about it.

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

New result:

$Prove using the Binomial Theorem or Combinatorics$

$\frac{1}{2} \binom{n}{0} + \frac{1}{3} \binom{n}{1} + \dots + \frac{1}{n+2} \binom{n}{n} = \frac{ 2^{n+1} \cdot n + 1}{(n+1)(n+2)}$

Note that the LHS is:

$\int_{0}^{1} x(1+x)^n \ dx$ (which we can derive easily)

Use u-substitution, we yield the RHS.

RealiseNothing · Mar 19, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$Find$ \ \ \ 100^2+99^2-98^2-97^2+96^2+95^2-94^2-93^2+ \dots + 4^2 + 3^2 -2^2 -1^2$

(The signs alternate in pairs)
(Provide proof of your value, either with latex or a brief outline of steps taken)

Consider them in groups of four. Now consider the general case:

$(x+2)^2 + (x+1)^2 - x^2 - (x-1)^2$

$x^2 + 4x + 4 + x^2 + 2x + 1 - x^2 - x^2 + 2x - 1$

$8x + 4$

$2(4x+2)$

Now we will note that:

$(x+2) + (x+1) + x + (x-1) = 4x+2$

So the sum of each group of four is just two times the sum of that group. Hence if we were to do it to all groups of four up to 100, we would get two times the sum of all the numbers from 1 to 100:

$2(1+2+3+...+98+99+100)$

This is just double the 100th triangular number, which is just $100\times101$

So the answer is $10100$

Sy123 · Mar 19, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
Consider them in groups of four. Now consider the general case:

$(x+2)^2 + (x+1)^2 - x^2 - (x-1)^2$

$x^2 + 4x + 4 + x^2 + 2x + 1 - x^2 - x^2 + 2x - 1$

$8x + 4$

$2(4x+2)$

Now we will note that:

$(x+2) + (x+1) + x + (x-1) = 4x+2$

So the sum of each group of four is just two times the sum of that group. Hence if we were to do it to all groups of four up to 100, we would get two times the sum of all the numbers from 1 to 100:

$2(1+2+3+...+98+99+100)$

This is just double the 100th triangular number, which is just $100\times101$

So the answer is $10100$

Elegant, nice work

Alternatively:

$100^2-98^2 + 99^2-97^2 + 96^2-94^2 + 95^2-93^2 + \dots + 4^2-2^2 + 3^2-1^2$

Difference of 2 squares

$(100-98)(100+98) + (99-97)(99+97) + \dots + (3-1)(3+1)$

Note that first bracket in each term is 2, take common:

$2(1+2+\dots + 100) = \cdot 100 \cdot 101 = 10100$

Sy123 · Mar 21, 2013

Re: HSC 2013 3U Marathon Thread

$$Find$ \ \ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{3^{m+n}}$

asianese · Mar 21, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$Find$ \ \ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{3^{m+n}}$

$\frac{9}{4}.$

Capt Rifle · Mar 22, 2013

Re: HSC 2013 3U Marathon Thread

<a href="http://www.codecogs.com/eqnedit.php?latex=Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" title="Find \int cos^42x\, sin^32x\, \, dx\, \: using\, the\, subsitution\, u=cos2x" /></a>

Edit: This is at a 3U difficulty, right guys?

Sy123 · Mar 22, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
$\frac{9}{4}.$

I don't believe you.

seanieg89 · Mar 22, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
I don't believe you.

I believe him.

Sy123 · Mar 22, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
I believe him.

I meant, prove your value :s

HSC 2013-14 MX1 Marathon (archive) (8 Viewers)

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