How to show $2\sum^{n/2}_{k=0}$ $(\frac{1}{2}-\frac{k}{n})\binom{n}{k} $ = $\frac{1}{2}$ $\binom{n}{n/2}$ How to show:
$$2\sum^{n/2}_{k=0}\left(\frac{1}{2}-\frac{k}{n}\right)\binom{n}{k}=\frac{1}{2}\binom{n}{n/2}$$
n:even
please could you help with this equality.
on page 17:  Rivlin, an intro to approximation of functions

Assuming you mean for $n$ to be even, write $n=2r$; then you want to show that
$$2\sum_{k=0}^r \left(\frac{1}{2}-\frac{k}{2r}\right)\dbinom{2r}{k} = \frac{1}{2}\dbinom{n}{n/2}.$$
Expanding the left-hand side gives
$$\sum_{k=0}^r \dbinom{2r}{k} - 2\sum_{k=0}^r \frac{k}{2r}\dbinom{2r}{k}.$$
The first term is the sum of the first $r+1$ terms in that row of Pascal's Triangle; it evaluates to
$$\frac{2^n-\binom{n}{n/2}}{2} + \dbinom{n}{n/2} = 2^{n-1} + \frac{1}{2}\dbinom{n}{n/2}.$$
(To see this, draw a picture of the row and figure out how to get the sum of those terms given that the sum of all terms in the row is $2^n$.)
The second term simplifies as
$$2\sum_{k=0}^r \frac{k}{2r}\cdot\frac{(2r)!}{k!(2r-k)!} = \sum_{k=1}^r \dbinom{2r-1}{k-1} = 2\sum_{k=0}^{r-1} \dbinom{2r-1}{k} = 2\cdot 2^{2r-2} = 2^{2r-1} = 2^{n-1}. $$
Putting the two together gives
$$2^{n-1} + \frac{1}{2}\dbinom{n}{n/2} - 2^{n-1} = \frac{1}{2}\dbinom{n}{n/2}.$$
