Summation with binomial coefficients question

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The discussion centers on evaluating the sum of binomial coefficients, specifically the expression \(\sum\limits_{r=0}^n\frac{1}{^nC_r}=a\) and the subsequent double sum involving indices \(i\) and \(j\). Participants note issues with consistency in variable substitution, particularly confusing \(m\) and \(n\) in their equations. There is a suggestion to focus on one term of the double sum to simplify the calculations. The challenge lies in correctly applying derived equations and ensuring proper substitutions to achieve the desired results. The conversation highlights the complexities involved in manipulating binomial coefficients and the need for careful attention to variable definitions.
AdityaDev
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Homework Statement


##\sum\limits_{r=0}^n\frac{1}{^nC_r}=a##. Then find the value of $$\sum\sum\limits_{0\le i<j\le n}(\frac{i}{^nC_i}+\frac{j}{^nC_j})$$

Homework Equations



I have used two equations which I derived myself. This is the first one.
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The second one is:
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3. The Attempt at a Solution

Using first equation and second equation:
20150430_001236-1.jpg

Now I have to subtract the cases where I=j to get the required sum. But Iis the above conclusion correct? Because I am not getting the required answer after subtracting.
 

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In the last line of the derivation of your second equation, you have been inconsistent in your substitutions of m for n. Two m's should be m+1.
But I don't understand how you use this equation anyway. a is a function of n, but where you use the equation you seem to be using it as a generic fact for any m, without changing a. I.e. you cannot now substitute n for m as being equal.
 
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Then how do you find ##(2n+1)\sum\limits_{r=0}^n\frac{r}{^nC_r}##?
 
AdityaDev said:
Then how do you find ##(2n+1)\sum\limits_{r=0}^n\frac{r}{^nC_r}##?
I haven't solved it myself, yet.
You could try concentrating on one of the two terms in the double sum. You should be able to sum that over the 'other' variable.
 

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