# Binomial Coefficients

by ritwik06
Tags: binomial, coefficients
 P: 586 1. The problem statement, all variables and given/known data If $$\sum^{n}_{r=0} \frac{1}{^{n}C_{r}} = a$$, then find the value of $$\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}$$ 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.
HW Helper
Thanks
PF Gold
P: 26,127
 Quote by ritwik06 1. The problem statement, all variables and given/known data If $$\sum^{n}_{r=0} \frac{1}{^{n}C_{r}} = a$$, then find the value of $$\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}$$ in terms of a and n.[/tex]
Hi ritwik06!

Hint: suppose n = 12.

Then $$\sum^{n}_{r=0} \frac{1}{^{n}C_{r}}$$

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

So what is $$\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}$$ ?
 HW Helper P: 2,618 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?
P: 586

## Binomial Coefficients

 Quote by tiny-tim Hi ritwik06! Hint: suppose n = 12. Then $$\sum^{n}_{r=0} \frac{1}{^{n}C_{r}}$$ = (0!12! + 1!11! + 2!10! + 3!9! + …)/12! So what is $$\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}$$ ?
Thank god! Somebody helped me. But Tim, I wonder what you wish to convey... Please could you be more explicit
 Sci Advisor HW Helper Thanks P: 24,990 Consider: $$\sum^{n}_{r=0} \frac{n-r}{^{n}C_{r}}$$ How does that compare with: $$\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}$$ Does that give you any ideas??
 Sci Advisor HW Helper Thanks PF Gold P: 26,127 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)
 Sci Advisor HW Helper Thanks P: 24,990 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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