<s>[INST] Proving that this sum $\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p^{2}$ How does one prove that for a prime $p \geq 5$ the sum :  $$\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$$ is divisible by $p^{2}$.
Since each term of $\displaystyle \sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p$, only thing remains is to prove that the sum $$\sum\limits_{ 0 < k < \frac{2p}{3}} \frac{1}{p} { p \choose k}$$ is divisible by $p$.
How to evaluate this sum: $\displaystyle \frac{1}{p} { p \choose k} = \frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdot 3 \cdots k}$
 [/INST] Since we are working in the field $\mathbb{F}_p$ we can write 
$$\frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdots k}$$ as
$$\frac{(-1)(-2) \cdots (-(k-1))}{1 \cdot2 \cdots k}$$
= $$\frac{(-1)^{k-1}}{k}$$
Let $N = [\frac{2p}{3}]$ and $M = [\frac{N}{2}]$
Thus what we need is
$$ \sum_{k=1}^{N} \frac{(-1)^{k-1}}{k}$$  $$  = \sum_{k=1}^{N} \frac{1}{k} - 2 \sum_{k=1}^{M}\frac{1}{2k}$$  $$ =\sum_{k=M+1}^{N}\frac{1}{k}$$
Now $N+M+1 =p$ so we can rewrite as 
$$\frac{1}{N} + \frac{1}{M+1} + \frac{1}{N-1} + \frac{1}{M+2} + \cdots = $$
$$\frac{p}{N(M+1)} + \frac{p}{(N-1)(M+2)} + \cdots $$
which is $0$.
There are $N-M$ terms, which is even, so each term gets paired off.
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