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Binomial Coefficients

  • Thread starter ritwik06
  • Start date
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1. Homework Statement
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.
 

tiny-tim

Science Advisor
Homework Helper
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1. Homework Statement
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! :smile:

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] ? :smile:
 

Defennder

Homework Helper
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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?
 
580
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Hi ritwik06! :smile:

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] ? :smile:
Thank god! Somebody helped me. But Tim, I wonder what you wish to convey... Please could you be more explicit :smile:
 

Dick

Science Advisor
Homework Helper
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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

Science Advisor
Homework Helper
25,790
249
Hi ritwik06!! :smile:

Have you got this now … you haven't said?

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

(same for the other thread)
 

Dick

Science Advisor
Homework Helper
26,258
618
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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