Summation with binomial coefficients question

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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.
Handwritten derivation expanding the double sum of (i + j) for i and j from 0 to n as a grid of terms, then regrouping it into (n+1) times the sum of r plus n times the sum of r, giving (2n+1) times the sum of r from r = 0 to n.

The second one is:
Handwritten derivation: sum from r=0 to n of 1/nCr equals 1 + sum from r=1 to n of r/(n-1)Cr-1 equals 1 + (1/n) times sum of r/(n-1)Cr-1 equals a; then sum from r=1 to n of r/(n-1)Cr-1 equals n(a-1); and for m = n-1, sum from r=1 to m of r/mCr equals m(a-1).


3. The Attempt at a Solution

Using first equation and second equation:
Handwritten derivation: sum over i and j from 0 to n of (i/nCi + j/nCj) equals (2n+1) times the sum over r from 0 to n of r/nCr, which equals (2n+1) times the sum from r=1 to n of r/nCr, equals (2n+1)n(a-1).

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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  • Handwritten derivation: sum from r=0 to n of 1/nCr equals 1 + sum from r=1 to n of r/(n-1)Cr-1 equals 1 + (1/n) times sum r/(n-1)Cr-1 equals a, then implies sum r/(n-1)Cr-1 = n(a-1), and for m = n-1, sum r/mCr = m(a-1).
    20150430_000435-1.jpg
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  • Handwritten derivation: sum from r=0 to n of 1/nCr equals 1 + sum from r=1 to n of r/(n-1)Cr-1 equals 1 + (1/n) times sum of r/(n-1)Cr-1 equals a, implying sum from r=1 to n of r/(n-1)Cr-1 equals n(a-1), and for m = n-1 the sum from r=1 to m of r/mCr equals m(a-1).
    20150430_000435-1.jpg
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  • Handwritten derivation of the double sum of (i + j) for i and j from 0 to n, expanded as a grid of terms, grouped by row and simplified to (2n + 1) times the sum of r from r = 0 to n.
    20150430_000946-1.jpg
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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.