Which term in the expansion of $(1+\sqrt{3})^{100}$ by Newton's binomial formula will be the largest?
Let's compare two consecutive terms: $\frac{C_{100}^{n+1}(\sqrt{3})^{n+1}}{C_{100}^{n}(\sqrt{3})^{n}}=\frac{100-n}{n+1} \cdot \sqrt{3}$. This number is greater than one when

$(100-n) \sqrt{3}>n+1$, which is when $n<\frac{100 \sqrt{3}-1}{\sqrt{3}+1}=63.03 \ldots$ Therefore, $C_{100}^{63}(\sqrt{3})^{63}$ is still smaller than the next term $C_{100}^{64}(\sqrt{3})^{64}$, and subsequently, the terms monotonically decrease.

## Answer

$C_{100}^{64}(\sqrt{3})^{64}$.