\displaystyle \sum_{k=0}^{1010}(-1)^k\dbinom{2021}{2k}= _______．
答案 -2^{1010}．
解析 根据题意, 有\begin{split}\displaystyle \sum_{k=0}^{1010}(-1)^k\dbinom{2021}{2k}&=\sum_{k=0}^{1010}{\rm i}^{2k}\dbinom{2021}{2k}\\ &=\mathop{\rm Re}(1+{\rm i})^{2021}\\ &=\mathop{\rm Re}\left(\dfrac{\pi}4:\sqrt 2\right)^{2021}\\ &=\mathop{\rm Re}\left(\dfrac{5\pi}4:2^{\frac{2021}2}\right)\\ &=-2^{1010}.\end{split}