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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

In class we studied the identity \displaystyle\binom{r}{r}+\binom{r+1}{r} +\binom{r+2}{r} + \cdots +\binom{n}{r} = \binom{n+1}{r+1} We also took a glimpse at \displaystyle\binom{r}{0}+\binom{r+1}{1} +\binom{r+2}{2} + \cdots +\binom{n}{n-r} = \binom{n+1}{n-r}. We will now take a closer look at this second identity.

(a) Confirm that the second identity works for n=5, r=2 and for n=7, r=3.

(b) What is the relationship between the first and second identities?

(c) Prove the second identity above algebraically without using what you learned in Part b. (In other words, prove it without the help of the hockey stick identity we studied in class).

(d) Prove the second identity above with a block-walking argument.

Genius is one percent inspiration and ninety-nine percent perspiration

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi;

Sorry, I can not read that?!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

srry, i used latex, i will change it to regular english

Genius is one percent inspiration and ninety-nine percent perspiration

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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

In class we studied the identity combination{r}{r}+combination{r+1}{r} +combination{r+2}{r} + ... +combination{n}{r} = combination{n+1}{r+1} We also took a glimpse at combination{r}{0}+combination{r+1}{1} +combination{r+2}{2} +... +combination{n}{n-r} = combination{n+1}{n-r}. We will now take a closer look at this second identity.

(a) Confirm that the second identity works for n=5, r=2 and for n=7, r=3.

(b) What is the relationship between the first and second identities?

(c) Prove the second identity above algebraically without using what you learned in Part b. (In other words, prove it without the help of the hockey stick identity we studied in class).

(d) Prove the second identity above with a block-walking argument.

Genius is one percent inspiration and ninety-nine percent perspiration

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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

combination{x}{y} means x combination y

Genius is one percent inspiration and ninety-nine percent perspiration

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

I know that.

combination{r}{r}+combination{r+1}{r} +combination{r+2}{r} + ... +combination{n}{r} = combination{n+1}{r+1} We also took a glimpse at combination{r}{0}+combination{r+1}{1} +combination{r+2}{2} +... +combination{n}{n-r} = combination{n+1}{n-r}.

It is totally unreadable on my browser

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

i will change it

Genius is one percent inspiration and ninety-nine percent perspiration

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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

In class we studied the identity (r) combination (r) + (r+1) combination (r) +(r+2) combination (r) + ... + (n) combination (r) = (n+1) combination r+1 We also took a glimpse at (r) combination (0) + (r+1) combination (1) +(r+2) combination (2) +... +(n) combination (n-r) = (n+1) combination (n-r). We will now take a closer look at this second identity.

(a) Confirm that the second identity works for n=5, r=2 and for n=7, r=3.

(b) What is the relationship between the first and second identities?

(c) Prove the second identity above algebraically without using what you learned in Part b. (In other words, prove it without the help of the hockey stick identity we studied in class).

(d) Prove the second identity above with a block-walking argument.

Genius is one percent inspiration and ninety-nine percent perspiration

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi;

What was your answer for a)?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

i didn't know how to do it

Genius is one percent inspiration and ninety-nine percent perspiration

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi;

Did you substitute the values in?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

i think i got it

Genius is one percent inspiration and ninety-nine percent perspiration

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Very good!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

never mind i didn't get it

Genius is one percent inspiration and ninety-nine percent perspiration

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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

how do you substitute it

Genius is one percent inspiration and ninety-nine percent perspiration

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

We also took a glimpse at (r) combination (0) + (r+1) combination (1) +(r+2) combination (2) +... +(n) combination (n-r) = (n+1) combination (n-r).

You want to prove this identity for n=5, r=2 and for n=7, r=3?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**mathstudent2000****Member**- Registered: 2013-07-26
- Posts: 79

yes please, but i don't know how to

Genius is one percent inspiration and ninety-nine percent perspiration

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

As near as I can understand it, for n=5, r=2.

We are done.

For n=7, r=3.

We are done.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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