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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

bobbym
2013-08-24 14:58:54

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





We are done.

For n=7, r=3.






We are done.

mathstudent2000
2013-08-24 13:44:21

yes please, but i don't know how to

bobbym
2013-08-22 10:45:17

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?

mathstudent2000
2013-08-22 10:18:08

how do you substitute it

mathstudent2000
2013-08-22 10:16:29

never mind i didn't get it

bobbym
2013-08-22 10:16:28

Very good!

mathstudent2000
2013-08-22 10:15:22

i think i got it

bobbym
2013-08-18 11:37:09

Hi;

Did you substitute the values in?

mathstudent2000
2013-08-18 09:41:29

i didn't know how to do it

bobbym
2013-08-18 02:55:32

Hi;

What was your answer for a)?

mathstudent2000
2013-08-18 02:51:53

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.

mathstudent2000
2013-08-18 02:45:44

i will change it

bobbym
2013-08-17 03:25:32

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

mathstudent2000
2013-08-17 03:17:51

combination{x}{y} means x combination y

mathstudent2000
2013-08-17 03:16:37

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.

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