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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).

 

Please just help with B!!! Step by step please

 Oct 7, 2017
 #1
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Jeff, why would I bother to help you when you are too rude to respond to my answers?

 

See here (just for an example)

 

https://web2.0calc.com/questions/probability-and-geometry#r7

 Oct 7, 2017
edited by Melody  Oct 7, 2017
 #17
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I AM SO SORRY MELODY!!! I never meant to be rude!!! I just didn't check my problems because I figured out before you helped me figure it out!!! I am super duper sorry! Please don't be mad at me...

Jeff123  Oct 13, 2017
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Melody, i dont think he meant to hurt you :)

 

Jeff123- you want to find the following sum:

 

(n choose 0)2+(n choose 1)2+........+(n choose n)2. Now, we need to find a clearer way to express this sum. As you can see, the sum involves around binomial coefficients (no shít blarney!). This is important, because there is a special way to solve this kind of question with a method called "the combinatorical proof"-

 

Suppose you have a sum that contains binomial coefficients, and suppose your teacher told you to find an easier way to express that sum. One way to solve this question (the combinatorical way) is to write a "story" that matches the sum- a story that involves around you having to choose cetrain things, where the number of ways you can choose the things is the sum, and where there is an EASIER way to count the ways you can choose (a clearer formula), and by that proving that the new formula you found finds the sum.

 

I'll give you an example- Suppose you want to find the sum (n choose 0)+(n choose 1)+........+(n choose n) (the sum of the nth row of pascal's triangle). Now, I'll write the story that matches the sum:

 

In bob's class (bob is the teacher) there are n students. He needs to choose some students that will clean the class (It is possible for him to choose NO students). What is the number of ways he choose the students?-

 

FIRST WAY TO DESCRIBE THE NUM' OF WAYS:

bob can choose 0 students OR 1 student OR 2 students......OR n students. He has exactly (n choose k) to choose k students, Therefore he has exactly (n choose 0)+(n choose 1)+........+(n choose n) ways to choose the students.

 

SECOND WAY TO DESCRIBE THE NUM' OF WAYS:

 

bob can grab his list of his students (in the list the students are numbered from 1 to n) and think to himself:

Do i want to choose the first student? 

Do i want to choose the second student?

Do i want to choose the third student?

.

.

.

Do i want to choose the nth student?

 

Everytime he asks himself that question, he has 2 options: the first option is to choose the student, and the second is to NOT choose him and let him leave the class. Therefore he has 2x2x2x2......x2 (n times)=2n ways to choose the students.

 

 

therefore 2n=(n choose 0)+(n choose 1)+........+(n choose n).

 

 

Can you do the same thing with the sum you gave us?

 

 

~blarney master~

 Oct 10, 2017
edited by Guest  Oct 10, 2017
 #3
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I never suggested that Jeff tried to hurt me!

 

He is just plain RUDE!  There is a huge difference.

Even when I pointed out his rudeness there was no suggestion of an explanation or an appology.

Perhaps is is because I am female?  

I see Jeff as an arrogant rude male who would not lower himself to thank a female for her help!

I think I have seen him be mildly polite to CPhill. This is why I think it is likely a sexist thing.

 

Jeff before you embarass yourself with a comment you have best know that the blarney master is female.

Melody  Oct 10, 2017
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Melody, i'm sorry for suggesting he hurted you. What i meant to say was that I think he never meant to do that out of rudeness (And i dont think he even knows that you're a female). But i'm sure he can apologise for it.

 

 

Also, i am not ginger ale, nor am i a female (DId you think i was ginger ale because of the way i wrote the word "shÍt"?)

 

 

~blarney master~

Guest Oct 10, 2017
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Well it is kinda strange how most of the time your English is perfect and then when you try to tell me you are not who I think you are that suddenly to write like English is your second language. 

Come on Ginger, you are smarter than that!

Melody  Oct 10, 2017
 #9
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GA a.k.a "JB" a.k.a "BM" a.k.a ~blarney master~ a.k.a "Devious" !!

Guest Oct 10, 2017
 #10
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W T F  Yes, Ginger is smarter than that! That is my proof these posts are not me. 

I recognize all the kooks on here.  Birds of a feather, for sure, but each has their own signature style. For god’s sake, no one needs to sign their name. I can smell who they are. 

GingerAle  Oct 10, 2017
 #11
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We can smell you too ginger ;)

 

And in response to your previous post about me being fùcked in the head: Doesn't my name imply that?

 

~blarney master a.k.a smelly kook~

Guest Oct 10, 2017
edited by Guest  Oct 10, 2017
 #13
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Yea, "Mind Reader Psycho !!".

Guest Oct 11, 2017
 #15
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What is your game?  I know BM is one of your puppets, TPM (The Puppet Master), but I don’t yet know what your game is. Is it some kind of self-styled therapy to deal with your hyperactive, chaotic brain? 

 

If so, please continue. Trolling you and the Blarney Banker is my self-styled cathartic therapy:  like squeezing zits, lancing boils, and breaking dams of batshit-stupid that keep the offal from flowing through the sewers of mystical Camelot.   

GingerAle  Oct 11, 2017
 #16
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Hail, hail the gang’s all here. I am amazed how the mere mention of my name brings in the ne’er-do-well pseudo intellectuals, the unsuccessful, the ineffectual, and the good for nothing –know-nothing dregs of mystical Camelot.

 

They gather like mutant vultures seeking revenge on me as if I am the cause of their brain-dead stupidity and virulent contagious dumbness.

 

You dimwits are so dumb to think smashing the fire alarm puts out the fire.indecision

GingerAle  Oct 11, 2017
 #8
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1^2 + 1^2  =  2  = 2 *C(1,1)

1^2 + 2^2 + 1^2  =  6      =   2 *C(3, 2)

1^2 + 3^2 + 3^2 + 1^2  = 20   =  2*C ( 5, 3)

1^2 + 4^2 + 6^2 + 4^2 + 1^2  = 70  = 2 * C (7,4)

1^2  + 5^2 + 10^2 + 10^2 + 5^2  + 1^2  = 252 =  2 * C ( 9, 5)  

 

The sum of the squares of the nth row entries in Pascal's Triangle seems to be :

 

2 * C ( 2n - 1 , n) 

 

 

Note that these successive sums appear as the "middle" term in row 2n of the triangle......!!!!

 

 

cool cool cool

 Oct 10, 2017
edited by CPhill  Oct 10, 2017
edited by CPhill  Oct 10, 2017
edited by CPhill  Oct 10, 2017
 #12
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2*C(2n-1, n)=C(2n, n). Yes, that is the answer. 


~blarney master~

Guest Oct 10, 2017
 #14
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BM This is an example of your disconnect to reality: Compare Sir CPhill’s answer to your meandering pile of incomprehensible slop. 

 

CPhill’s answer is compact, to the point, and clear. His comment is relevant and concise.

 

Your answer meanders all over Hades. By the time anyone wades though and digests this hog slop of discontinuity, the reader forgets the question, as the toxic brew induces a mental fog and accelerates brain atrophy.

GingerAle  Oct 11, 2017
 #18
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+1

Thanks everyone!!! Sorry for the late reply

 Oct 13, 2017

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