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Binomial Coefficients

by ritwik06
Tags: binomial, coefficients
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ritwik06
#1
Aug18-08, 02:42 PM
P: 582
1. The problem statement, all variables and given/known data
If [tex]\sum^{n}_{r=0} \frac{1}{^{n}C_{r}} = a[/tex], then find the value of [tex]\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}[/tex] in terms of a and n.[/tex]






3. The attempt at a solution
I tried to write down the terms of both the series, but to no avail. i cant think of anything.Please shed some light.
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tiny-tim
#2
Aug18-08, 06:38 PM
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Quote Quote by ritwik06 View Post
1. The problem statement, all variables and given/known data
If [tex]\sum^{n}_{r=0} \frac{1}{^{n}C_{r}} = a[/tex], then find the value of [tex]\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}[/tex] in terms of a and n.[/tex]
Hi ritwik06!

Hint: suppose n = 12.

Then [tex]\sum^{n}_{r=0} \frac{1}{^{n}C_{r}}[/tex]

= (0!12! + 1!11! + 2!10! + 3!9! + )/12!

So what is [tex]\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}[/tex] ?
Defennder
#3
Aug18-08, 09:52 PM
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Hi tim, I'm not seeing how this helps to solve the problem. You have a term dependent r in each summand, so how do we express it in a?

ritwik06
#4
Aug19-08, 12:59 PM
P: 582
Binomial Coefficients

Quote Quote by tiny-tim View Post
Hi ritwik06!

Hint: suppose n = 12.

Then [tex]\sum^{n}_{r=0} \frac{1}{^{n}C_{r}}[/tex]

= (0!12! + 1!11! + 2!10! + 3!9! + )/12!

So what is [tex]\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}[/tex] ?
Thank god! Somebody helped me. But Tim, I wonder what you wish to convey... Please could you be more explicit
Dick
#5
Aug19-08, 02:59 PM
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Consider:
[tex]
\sum^{n}_{r=0} \frac{n-r}{^{n}C_{r}}
[/tex]
How does that compare with:
[tex]
\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}
[/tex]
Does that give you any ideas??
tiny-tim
#6
Aug22-08, 03:44 PM
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Hi ritwik06!!

Have you got this now you haven't said?

If you haven't, then follow Dick's hint it's much better than mine!

(same for the other thread)
Dick
#7
Aug23-08, 11:51 PM
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That's nice of you to say, tiny-tim. Thanks. :) Now you've got me curious. ritwik06, did you get it? It's surprising easy if you think about it right, and pretty nonobvious if you don't. It took me a while.


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