+1836 $$In class we saw that the sum of the entries of row $n$ of Pascal's Triangle is $2^n$. In this problem we investigate the sums of the squares of the entries of row $n$ of Pascal's Triangle.
(a) Compute the sums of the squares of Rows 1-4 of Pascal's Triangle. That is, compute:
$$\binom10^2 + \binom11^2$$
$$\binom20^2 + \binom21^2 + \binom22^2$$
$$\binom30^2 + \binom31^2 + \binom32^2 + \binom33^2$$
$$\binom40^2 + \binom41^2 + \binom42^2 + \binom43^2 + \binom44^2$$
Do these sums appear anywhere else in Pascal's Triangle?
(b) Guess at an identity based on your observations from part (a). Your identity should be of the form
$$\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \text{ something}.$$
(You have to figure out what "something" is.) Test your identity for $n=1,2,3,4$ using your results from part (a).
(c) Prove your identity using a committee-forming argument.
(d) Prove your identity using a block-walking argument.$$
Please explain very well for each part in this question.
+129852 Here's part of the question.......
a) The sums are
2 = C(2,1)
6 = C(4, 2)
20 = C(6, 3)
70= C(8,4)
b) The pattern appears to be
[C(n,0)]^2 + [C(n, 1)]^2 + [C(n, 2)]^2 +.....+ [C(n,n)]^2 = C(2n, n)
Sorry....I don't know the answers to c or d !!!!
+118673 I am sorry Mellie I don't know either. I don't even know what committee forming arguements or block walking arguments are :/ 
I'll put it in the wrap, maybe one of the other mathematicians can help you 
Imagine Pascal's Triangle with lines connecting each term. The ways go from the top to somewhere else is a block-walking proof in this case.
As for the committee-forming proof, we need to form a committee by using
\(\dbinom{n}{k}\)
Committee forming is just like saying you have $n$ people, and from that $n$ people, you want to choose $k$ people to form a committee or group, so how many different groups can you make? So it's just like $\binom{n}{k}$.
