Calculating Sum of Binomial Coefficients in Terms of a and n

[ritwik06]

Homework Statement

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]

The Attempt at a Solution

I tried to write down the terms of both the series, but to no avail. i can't think of anything.Please shed some light.

 

[tiny-tim]

Homework Statement

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}} ?

 

[Defennder]

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]

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

 

[Dick]

Consider:
<br /> \sum^{n}_{r=0} \frac{n-r}{^{n}C_{r}}<br />
How does that compare with:
<br /> \sum^{n}_{r=0} \frac{r}{^{n}C_{r}}<br />
Does that give you any ideas??

 

[tiny-tim]

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]

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.