+118723 Hi Itzcubez,
I have no idea what c or d even mean. Sorry. Here's the first bit.
(a) Compute the sums of the squares of Rows 1-4 of Pascal's Triangle.
\(1+1=2\\ 1+4+1=6\\ 1+9+9+1=20\\ 1+16+36+16+1=70\\ \binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \frac{(2n)!}{(n!)^2} \)
Do these sums appear anywhere else in Pascal's Triangle?
\( \binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \frac{(2n)!}{(n!)^2}\)
(b) Guess at an identity based on your observations from part (a). Your identity should be of the form
(You have to figure out what "something" is.) Test your identity for using your results from part (a).
You can do the testing but yea it works :)
(c) Prove your identity using a committee-forming argument.
(d) Prove your identity using a block-walking argument.
This is a repost of http://web2.0calc.com/questions/in-class-we-saw-that-the-sum-of-the-entries-of-row-n-of-pascal-s-triangle-is-2-n-in-this-problem-we-investigate-the-sums-of-the-squares which is totally dead.
Source: AoPS
If you are going to answer please provide your reasoning for Parts (c) and (d)
Oh I see this is one of Mellie's old questions - She asked some great questions :)
+118723 LaTex discussion on displaying fractions :)
Heureka, I thought maybe you would like to add to this post :/ 
I have also been looking at the Latex that is used in this question.
I found this reference to using Latex for fractions but i will admit i have not properly digested it yet.
http://tex.stackexchange.com/questions/59747/proper-display-of-fractions
This is the question:
\(\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\)
And this is the command fot it.
\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}
I have not used \cfrac{}{} before.
On that reference site there is also a \dfrac{}{} used.
If anyone would like to discuss how these are used that would because I haven't fully comprehended what they are about.
+118723 Hi Alan
I have been toying this question, I have only just worked out the relevance of the 2 answers :D
Silly me :)
Here is Alan's earlier answer:
http://web2.0calc.com/questions/find-the-value-of_3
INFINITE CONTINUED FRACTION
\( \begin{align} Let\\ x&=\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\\ so\\ x&=\cfrac{1}{1 + \cfrac{1}{2 +x}}\\~\\ x&=\cfrac{1}{\cfrac{2+x}{2+x} + \cfrac{1}{2 +x}}\\~\\ x&=\cfrac{1}{\frac{3+x}{2+x} }\\~\\ x&=\cfrac{2+x}{3+x}\\~\\ 3x+x^2&=2+x\\ x^2+2x-2&=0\\ x&=\frac{-2\pm\sqrt{4+8}}{2}\\ x&=\frac{-2\pm2\sqrt{3}}{2}\\ x&=-1\pm\sqrt{3}\\ \end{align}\)
BUT it can be seen that x is positive so the only valid solutions is \(x=\sqrt3-1\)
SO
\(\boxed{\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\\~\\ =\sqrt3-1}\)
WHEREAS IF IT IS NOT AN INFINITE CONTINUED FRACTION THEN
\(\boxed{\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 }}}}\\~\\ =\frac{8}{11}}\)
.
