Binomial Coefficients

1. ritwik06

582
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.

2. tiny-tim

26,054
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 $$\sum^{n}_{r=0} \frac{r}{^{n}C_{r}}$$ ?

3. Defennder

2,616
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?

4. ritwik06

582
Thank god! Somebody helped me. But Tim, I wonder what you wish to convey... Please could you be more explicit

5. Dick

25,663
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??

6. tiny-tim

26,054
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)

7. Dick

25,663
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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