Let $p$ be an odd prime and let $n$ be a positive integer. For any positive integer $α$ and $m\in\{1,2,3\}$, we have \begin{align*} \sum_{k=0}^{p^αn-1}\frac{(\frac12)_k}{k!}\cdot\frac{(-4)^k}{m^k}\equiv\bigg(\frac{m(m-4)}{p}\bigg)\sum_{k=0}^{p^{α-1}n-1}\frac{(\frac12)_k}{k!}\cdot\frac{(-4)^k}{m^k}\pmod{p^{2α}}, \end{align*} where $(x)_k=x(x+1)\cdots(x+k-1)$ and $\big(\frac{\cdot}{\cdot}\big)$ denotes the Legendre symbol. Also, when $m=4$, \begin{align*} \sum_{k=0}^{p^αn-1}(-1)^k\cdot\frac{(\frac12)_k}{k!}\equiv p\sum_{k=0}^{p^{α-1}n-1}(-1)^k\cdot\frac{(\frac12)_k}{k!}\pmod{p^{2α}}. \end{align*}