New congruences for central binomial coefficients

Let $p$ be a prime, and let $d\in\{0,...,p^a\}$ with $a\in\Z^+$. In this paper we determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=1}^{p-1}\binom{2k}{k+d}/(km^{k-1})$ modulo $p$ where $m$ is an integer not divisible by $p$. For example, we show that if $p\not=2,5$ then $$\sum_{k=1}^{p-1}(-1)^k\frac{\binom{2k}k}k=-5\frac{F_{p-(\frac p5)}}p (mod p),$$ where $F_n$ denotes the $n$th Fibonacci number. We also prove that if $p>3$ then $$\sum_{k=1}^{p-1}\frac{\binom{2k}k}k={8/9} p^2B_{p-3} (mod p^3),$$ where $B_n$ is the $n$th Bernoulli number.