根据题意有$$f^2(n+1)-f(n+1)+f^2(n)-f(n)=-\dfrac14,$$即$$a_{n+1}+a_n=-\dfrac14,n\in\mathbb N^\ast,$$因此有$$a_{n}=a_{n+2},n\in\mathbb N^\ast,$$于是$$\begin{split} \sum_{k=1}^{15}a_k=8a_1+7a_2,\end{split}$$又结合$$\left(\sum_{k=1}^{15}a_k=-\dfrac{31}{16}\right)\wedge\left(a_1+a_2=-\dfrac14\right),$$于是解得$$(a_1,a_2)=\left(-\dfrac3{16},-\dfrac1{16}\right),$$显然$$\forall x\in\mathbb R,f(x+1)\geqslant \dfrac12,$$故$$f(n)=\dfrac12+\sqrt{a_n+\dfrac14},n\in\mathbb N^\ast,$$因此$$\begin{split}
\sum_{k=1}^{1000}f(k)&=500\left(f(1)+f(2)\right)\\
&=500+500\left(\sqrt{a_1+\dfrac14}+\sqrt{a_2+\dfrac14}\right)\\
&=625+125\sqrt3.\end{split}$$