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Binomial Coefficients 
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#1
Aug1808, 02:42 PM

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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. 


#2
Aug1808, 06:38 PM

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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] ? 


#3
Aug1808, 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?



#4
Aug1908, 12:59 PM

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Binomial Coefficients



#5
Aug1908, 02:59 PM

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Consider:
[tex] \sum^{n}_{r=0} \frac{nr}{^{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?? 


#6
Aug2208, 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) 


#7
Aug2308, 11:51 PM

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That's nice of you to say, tinytim. 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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