$$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.

Mellie
Apr 24, 2015

#1**+16 **

Best Answer

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 !!!!

CPhill
Apr 25, 2015

#3**0 **

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

Melody
Apr 28, 2015

#4**0 **

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

\(\dbinom{n}{k}\)

Guest Apr 24, 2016

#5**0 **

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 $\binom{n}{k}$.

Guest Jul 26, 2016