Binomial coefficient

[Trenters4325]

Prove that:

\(\displaystyle \L \, 2\, \sum^{[\frac{n\, -\, 1}{2}]}_{r=0}\{\frac{n\, -\, 2r}{n}\, \left(C(n,r)\right)\}^2\, =\, \sum^{n}_{r=0} \left{ \, \frac{n\, -\, 2r}{n}\, \left(C(n,r)\right)\ \, \right}^2\)

where C is the binomial coefficient and [] is the greatest integer function.

The substitution s = n - r should be useful.

 

[Deleted member 4993]

Please show us your work - indicating exactly where you are stuck - then we can provide useful help.

 

[Opalg]

Notice that the formula for the r'th term of the series is exactly the same on the two sides of this identity. Start by looking at the series on the right-hand side. In this series, the r'th term is the same as the (n-r)'th term (that's what the "useful" hint is trying to tell you - check it out!). So to get the sum of the whole series, you can just take the sum of the first half of the terms and then multiply the result by 2. There's a little complication if n is even, because then there is an odd number of terms. But in that case the middle term vanishes because of the factor n-2r. Conclusion: the sum on the right-hand side of the identity is twice the sum on the left-hand side.