3. In the expansion of $(\sqrt{3}+i)^{10}$, the sum of all odd terms is $\qquad$ .
3. 512 .

It is known that in the expansion of $(\sqrt{3}+\mathrm{i})^{10}$, the sum of all odd terms is the real part of the complex number.
$$
\begin{array}{l}
\text { Therefore, }(\sqrt{3}+i)^{10}=\left((-2 i)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)\right)^{10} \\
=(-2 i)^{10}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{10} \\
=(-1024)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right) \\
=512-512 \sqrt{3} \mathrm{i}
\end{array}
$$