Primes dividing sum of binomial coefficients. Given that $p>3$ is a prime, we have $k=\lfloor \frac {2p}{3} \rfloor$ then prove or disprove that $\sum_{i=1}^k\binom{p}{i}\equiv 0\pmod{p^2}$.
Its easy to see how $p$ divides each of the binomial coefficients, but I tried using maximal powers of $p$ in each coefficient but was stuck can someone help.
 A: We want to prove that $\sum \limits_{i=1}^{\lfloor \frac{2p}{3} \rfloor} \binom{p}{i} = 0 \mod p^2$ for $p$- primes.
$\sum \limits_{i=1}^{\lfloor \frac{2p}{3} \rfloor} \binom{p}{i} = 0 \mod p^2$ imply that $\sum \limits_{i=1}^{\lfloor \frac{2p}{3} \rfloor} \frac{1}{p}\binom{p}{i}  =  \sum \limits_{i=1}^{\lfloor \frac{2p}{3} \rfloor} \frac{(-1)^{i-1}}{i} = 0 \mod p $ (as mentioned in the comments).
For $p=3k+1$ => $\sum \limits_{i=1}^{2k} \frac{(-1)^{i-1}}{i} = 1+\sum \limits_{i=1}^{k-1}\frac{1}{2i+1}-\sum \limits_{i=1}^{k} \frac{1}{2i} = \frac{1}{2} \left(H_{k-\frac{1}{2}}-H_k+\log (4)\right)$ 
and since $H_{2k}=\frac{1}{2}(H_k+H_{k-1/2}+\ln 4)$ 
we can get that 
$\frac{1}{2}(H_{k-1/2}-H_k+\ln 4)-\frac{1}{2}(H_{k-1/2}+H_k+\ln 4) = -H_{2k} \mod (3k+1)$.
Which is just $-H_k = -H_{2k} \mod (3k+1)$ 
Rearranging that we arrive at $H_{2k}-H_k = 0 \mod (3k+1)$ when $p=3k+1$ is prime.
For $p=3k+2$ => $\sum \limits_{i=1}^{2k+1} \frac{(-1)^{i-1}}{i} = 1+\sum \limits_{i=1}^{k-1}\frac{1}{2i+1}-\sum \limits_{i=1}^{k} \frac{1}{2i}+\frac{1}{2k+1} = \frac{1}{2} \left(H_{k-\frac{1}{2}}-H_k+\log (4)\right)+\frac{1}{2k+1}$ 
and since $H_{2k}=\frac{1}{2}(H_k+H_{k-1/2}+\ln 4)$ 
we can get that 
$\frac{1}{2}(H_{k-1/2}-H_k+\ln 4)-\frac{1}{2}(H_{k-1/2}+H_k+\ln 4)+\frac{1}{2k+1} = -H_{2k} \mod (3k+2)$.
Which is just $-H_k+\frac{1}{2k+1} = -H_{2k} \mod (3k+2)$ 
Rearranging that we arrive at $H_{2k}-H_k+\frac{1}{2k+1} = 0 \mod (3k+2)$ when $p=3k+2$ is prime.
Note : $H_k = \sum \limits_{i=1}^{k} \frac{1}{i}$ is the Harmonic number and also the highlighted part,are the statements that i did not prove.
A: Ahmad showed the problem is equivalent to


*

*$H_{2k}-H_k \equiv 0 \pmod p$ for $p \equiv 1 \pmod 3$

*$H_{2k}-H_k+(2k+1)^{-1}  \equiv 0 \pmod p$ for $p \equiv 2 \pmod 3$


1:
$$H_{2k}-H_k \equiv \sum_{i=1}^{2k} i^{-1} - \sum_{i=1}^{k} i^{-1} \equiv \sum_{i=k+1}^{2k} i^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=\frac{3}{2}k+1}^{2k} i^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=-\frac{3}{2}k-1}^{-2k} (-i)^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=3k+1-\frac{3}{2}k-1}^{3k+1-2k} (-i)^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=k+1}^{\frac{3}{2}k} (-i)^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=k+1}^{\frac{3}{2}k} (-1)^{-1}(i)^{-1}\equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + -i^{-1}
\equiv
\sum_{i=k+1}^{\frac{3}{2}k} 0
\equiv
0
\pmod p$$
The idea is that for each inverse the opposite inverse is also in the sum and therefore the sum cancelled out. You need to show $k$ is even so no inverse has no companion. If $k$ is odd, $3k+1$ is divisible by $2$ and therefore not a prime.
2:
Same idea just with odd number of terms.
$$\begin{align}
H_{2k}-H_k+(2k+1)^{-1}  \equiv \\ &
\equiv
(2k)^{-1}+(2k-1)^{-1}+\cdots+(1)^{-1} - ((k)^{-1}+(k-1)^{-1}+\cdots+(1)^{-1}) + (2k+1)^{-1} \\ &
\equiv
(2k)^{-1}+(2k-1)^{-1}+\cdots+(k+1)^{-1} +(2k+1)^{-1} \\ &
\equiv
(2k)^{-1}+(2k-1)^{-1}+\cdots+(k+1)^{-1} + (2k+1)^{-1} \\ &
\equiv
-(k+2)^{-1}+-(k+3)^{-1}+\cdots+(k+1)^{-1} + -(k+1)^{-1} \\ &
\equiv
0
\pmod p \end{align}$$
Again we need to show $k$ is always odd. If $k$ is even $3k+2$ is divisible by $2$.
