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.

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.
